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7e113d0340
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581d26b018
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#include <stdio.h> |
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#include "utils.h" |
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#include <stdlib.h> |
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/*
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We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once; for example, the 5-digit number, 15234, is 1 through 5 pandigital. |
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The product 7254 is unusual, as the identity, 39 × 186 = 7254, containing multiplicand, multiplier, and product is 1 through 9 pandigital. |
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Find the sum of all products whose multiplicand/multiplier/product identity can be written as a 1 through 9 pandigital. |
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HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum. |
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https://projecteuler.net/problem=32
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*/ |
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int main(int argc,char**argv) { |
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int numb1=1; |
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int numb2=1; |
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long sumPandigital=0; |
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boolean*uniqueProducts = malloc(100000000); |
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for (int i=0;i<100000000;i++) { |
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uniqueProducts[i]=false; |
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} |
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for (numb1=1;numb1<10000;numb1++) { |
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for (numb2=1;numb2<10000;numb2++) { |
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int prod=numb1*numb2; |
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int numb=prod; |
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boolean uniqueDigits[9] = {}; |
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while (numb>0) { |
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int digit=numb%10-1; |
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if (digit!=-1&&!uniqueDigits[digit]) { |
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uniqueDigits[digit]=true; |
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} else { |
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goto skip; |
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} |
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numb/=10; |
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} |
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numb=numb1; |
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while (numb>0) { |
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int digit=numb%10-1; |
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if (digit!=-1&&!uniqueDigits[digit]) { |
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uniqueDigits[digit]=true; |
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} else { |
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goto skip; |
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} |
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numb/=10; |
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} |
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numb=numb2; |
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while (numb>0) { |
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int digit=numb%10-1; |
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if (digit!=-1&&!uniqueDigits[digit]) { |
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uniqueDigits[digit]=true; |
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} else { |
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goto skip; |
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} |
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numb/=10; |
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} |
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for (int i=0;i<9;i++) { |
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if (!uniqueDigits[i]) { |
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goto skip; |
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} |
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} |
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//If we got here it's a pandigital.
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if (!uniqueProducts[prod]) { |
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uniqueProducts[prod]=true; |
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sumPandigital+=prod; |
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printf("\n%d*%d = %d is pandigital!",numb1,numb2,prod); |
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} |
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skip: |
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continue; |
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} |
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} |
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printf("\nSum of pandigitals: %ld",sumPandigital); |
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return 0; |
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} |
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#include "utils.h" |
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#include <stdio.h> |
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#include <stdlib.h> |
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struct String mult(struct String numb1, struct String numb2) { |
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struct String n1 = numb1; |
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struct String n2 = numb2; |
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byte carryover = 0; |
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if (numb2.length>numb1.length) { |
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n1=numb2; |
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n2=numb1; |
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} |
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int addends[n2.length][n1.length+1]; |
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for (int i=0;i<n2.length;i++) { |
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for (int j=0;j<n1.length+1;j++) { |
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addends[i][j]=0; |
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} |
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} |
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for (int i=n2.length-1;i>=0;i--) { |
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carryover=0; |
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for (int j=n1.length-1;j>=0;j--) { |
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int mult = (n1.str[j]-'0')*(n2.str[i]-'0')+((carryover!=0)?carryover:0); |
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//printf(" %d/%d\n",mult,carryover);
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carryover=0; |
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if (mult>=10) { |
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carryover=mult/10; |
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mult=mult%10; |
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} |
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addends[(n2.length-1)-i][j+1]=mult; |
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} |
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if (carryover>0) { |
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addends[(n2.length-1)-i][0]=carryover; |
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} |
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} |
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//printIntDoubleArr(n2.length,n1.length+1,addends);
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struct String sum = {1,"0"}; |
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for (int i=0;i<n2.length;i++) { |
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char val[n1.length+1+i]; |
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for (int j=0;j<n1.length+1+i;j++) { |
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val[j]='0'; |
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} |
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for (int j=0;j<n1.length+1;j++) { |
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val[j]=addends[i][j]+'0'; |
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} |
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sum=add((struct String){n1.length+1+i,val},sum); |
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//printf("\nAA:%s",sum.str);
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} |
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if (sum.str[0]=='0') { |
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char*newStr=malloc(sum.length); |
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for (int i=1;i<sum.length+1;i++) { |
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newStr[i-1]=sum.str[i]; |
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} |
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free(sum.str); |
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sum=(struct String){sum.length-1,newStr}; |
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} |
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//printf("\nA:%s",sum.str);
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return sum; |
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} |
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struct String add(struct String numb1, struct String numb2){ |
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//printf("%s %s\n",numb1.str,numb2.str);
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byte carryover=0; |
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int digitCount=0; |
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char*str = malloc(1); |
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//str[0]='\0';
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if (numb1.length>=numb2.length) { |
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for (int offset=0;offset<numb1.length;offset++) { |
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str = realloc(str,++digitCount); |
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//printf("Digit count is now %d\n",digitCount);
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if (numb2.length>offset) { |
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//printf("%c %c\n",numb1.str[numb1.length-offset-1],numb2.str[numb2.length-offset-1]);
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int sum=((numb1.str[numb1.length-offset-1]-'0')+(numb2.str[numb2.length-offset-1]-'0'))+((carryover>0)?carryover--:0); |
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if (sum>=10) { |
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carryover=1; |
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sum-=10; |
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} |
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str[offset]=sum+'0'; |
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} else { |
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str[offset]=numb1.str[numb1.length-offset-1]+((carryover>0)?carryover--:0); |
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} |
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//str[offset+1]='\0';
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} |
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} else { |
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for (int offset=0;offset<numb2.length;offset++) { |
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str = realloc(str,++digitCount); |
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//printf("Digit count is now %d\n",digitCount);
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if (numb1.length>offset) { |
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//printf("%c %c\n",numb1.str[numb1.length-offset-1],numb2.str[numb2.length-offset-1]);
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int sum=((numb1.str[numb1.length-offset-1]-'0')+(numb2.str[numb2.length-offset-1]-'0'))+((carryover>0)?carryover--:0); |
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if (sum>=10) { |
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carryover=1; |
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sum-=10; |
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} |
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str[offset]=sum+'0'; |
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} else { |
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str[offset]=numb2.str[numb2.length-offset-1]+((carryover>0)?carryover--:0); |
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} |
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//str[offset+1]='\0';
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} |
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} |
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if (carryover>0) { |
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str = realloc(str,++digitCount); |
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str[digitCount-1]='1'; |
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//str[digitCount]='\0';
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} |
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for (int i=0;i<digitCount/2;i++) { |
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char c = str[i]; |
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char c2 = str[digitCount-i-1]; |
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str[digitCount-i-1]=c; |
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str[i]=c2; |
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} |
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str = realloc(str,digitCount+1); |
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str[digitCount]='\0'; |
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struct String newStr = {digitCount,str}; |
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return newStr; |
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} |
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void printLongDoubleArr(int a,int b,long doubleArr[a][b]) { |
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for (int i=0;i<a;i++) { |
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for (int j=0;j<b;j++) { |
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printf("%ld\t",doubleArr[i][j]); |
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} |
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printf("\n"); |
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} |
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} |
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void printIntDoubleArr(int a,int b,int doubleArr[a][b]) { |
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for (int i=0;i<a;i++) { |
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for (int j=0;j<b;j++) { |
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printf("%d\t",doubleArr[i][j]); |
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} |
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printf("\n"); |
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} |
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} |
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struct String createBigNumber(char*numb) { |
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int marker=0; |
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while (numb[marker++]!='\0'); |
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return (struct String){marker-1,numb}; |
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} |
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int*getFactors(int numb) { |
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int*factorList=malloc(sizeof(int)*1); |
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int factorListSize=2; |
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factorList[0]=1; |
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factorList[1]=numb; |
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int max=numb; |
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for (int i=2;i<max;i++) { |
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if (numb%i==0) { |
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factorList=realloc(factorList,sizeof(int)*++factorListSize); |
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factorList[factorListSize-1]=i; |
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if (numb/i!=i) { |
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factorList=realloc(factorList,sizeof(int)*++factorListSize); |
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factorList[factorListSize-1]=numb/i; |
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max=numb/i; |
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} |
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} |
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} |
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factorList=realloc(factorList,sizeof(int)*++factorListSize); |
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factorList[factorListSize-1]=0; |
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return factorList; |
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} |
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boolean isFactor(int*factorList,int numb) { |
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int counter=0; |
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while (factorList[counter]!=0) { |
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if (numb==factorList[counter]) { |
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return true; |
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} |
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counter++; |
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} |
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return false; |
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} |
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struct String BigPow(int numb1, int numb2) { |
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struct String sum = BigNumber(1); |
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char*newStr = malloc(0); |
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int strLength=0; |
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while (numb1>0) { |
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newStr=realloc(newStr,++strLength); |
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newStr[strLength-1]=(numb1%10)+'0'; |
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numb1/=10; |
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} |
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for (int i=0;i<strLength/2;i++) { |
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char oldC=newStr[strLength-i-1]; |
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newStr[strLength-i-1]=newStr[i]; |
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newStr[i]=oldC; |
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} |
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struct String bigNumb2 = (struct String){strLength,newStr}; |
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for (int i=0;i<numb2;i++) { |
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sum=mult(sum,bigNumb2); |
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} |
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free(newStr); |
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return sum; |
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} |
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boolean equals(struct String str1,struct String str2) { |
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if (str1.length!=str2.length) { |
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return false; |
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} |
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for (int i=0;i<str1.length;i++) { |
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if (str1.str[i]!=str2.str[i]) { |
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return false; |
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} |
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} |
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return true; |
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} |
@ -0,0 +1,18 @@ |
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#define true 1 |
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#define false 0 |
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#define boolean char |
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#define byte char |
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struct String{ |
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int length; |
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char*str; |
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}; |
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struct String add(struct String numb1, struct String numb2); |
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struct String mult(struct String numb1, struct String numb2); |
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struct String BigPow(int numb1,int numb2); |
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void printLongDoubleArr(int a,int b,long doubleArr[a][b]); |
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void printIntDoubleArr(int a,int b,int doubleArr[a][b]); |
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struct String createBigNumber(char*numb); |
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#define BigNumber(X) createBigNumber(#X) |
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boolean isFactor(int*factorList,int numb); |
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int*getFactors(int numb); |
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boolean equals(struct String str1,struct String str2); |
@ -1,99 +1,76 @@ |
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#include <stdio.h> |
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#include "utils.h" |
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#include <stdlib.h> |
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/*
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In the United Kingdom the currency is made up of pound (£) and pence (p). There are eight coins in general circulation: |
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We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once; for example, the 5-digit number, 15234, is 1 through 5 pandigital. |
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1p, 2p, 5p, 10p, 20p, 50p, £1 (100p), and £2 (200p). |
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The product 7254 is unusual, as the identity, 39 × 186 = 7254, containing multiplicand, multiplier, and product is 1 through 9 pandigital. |
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It is possible to make £2 in the following way: |
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Find the sum of all products whose multiplicand/multiplier/product identity can be written as a 1 through 9 pandigital. |
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1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p |
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HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum. |
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How many different ways can £2 be made using any number of coins? |
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https://projecteuler.net/problem=31
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https://projecteuler.net/problem=32
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*/ |
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enum Currency{ |
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pence1, |
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pence2, |
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pence5, |
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pence10, |
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pence20, |
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pence50, |
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pound1, //100 pences
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pound2, //200 pences
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}; |
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int getValue(enum Currency cur) { |
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switch (cur) { |
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case pence1:{ |
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return 1; |
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}break; |
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case pence2:{ |
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return 2; |
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}break; |
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case pence5:{ |
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return 5; |
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}break; |
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case pence10:{ |
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return 10; |
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}break; |
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case pence20:{ |
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return 20; |
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}break; |
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case pence50:{ |
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return 50; |
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}break; |
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case pound1:{ |
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return 100; |
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}break; |
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case pound2:{ |
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return 200; |
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}break; |
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int main(int argc,char**argv) { |
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int numb1=1; |
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int numb2=1; |
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long sumPandigital=0; |
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boolean*uniqueProducts = malloc(100000000); |
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for (int i=0;i<100000000;i++) { |
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uniqueProducts[i]=false; |
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} |
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} |
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int total(int*amts) { |
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return
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getValue(pence1)*amts[0]+ |
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getValue(pence2)*amts[1]+ |
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getValue(pence5)*amts[2]+ |
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getValue(pence10)*amts[3]+ |
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getValue(pence20)*amts[4]+ |
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getValue(pence50)*amts[5]+ |
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getValue(pound1)*amts[6]+ |
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getValue(pound2)*amts[7]; |
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} |
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void carryover(int*amts,int placeVal) { |
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amts[placeVal]++; |
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if (placeVal<7&&amts[placeVal]>200/getValue(placeVal)) { |
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carryover(amts,placeVal+1); |
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amts[placeVal]=0; |
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for (numb1=1;numb1<10000;numb1++) { |
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for (numb2=1;numb2<10000;numb2++) { |
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int prod=numb1*numb2; |
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int numb=prod; |
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boolean uniqueDigits[9] = {}; |
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while (numb>0) { |
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int digit=numb%10-1; |
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if (digit!=-1&&!uniqueDigits[digit]) { |
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uniqueDigits[digit]=true; |
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} else { |
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goto skip; |
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} |
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} |
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int main(int argc,char**argv) { |
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int currencyAmts[8] = {}; |
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int combinationCount=0; |
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int maxCurrencyVal=200; |
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int currentMarker=0; |
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while (currencyAmts[6]!=2) { |
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if (total(currencyAmts)==200) { |
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printf("Coin amts: "); |
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for (int i=0;i<8;i++) { |
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printf("%d ",currencyAmts[i]); |
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numb/=10; |
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} |
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printf("\n"); |
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combinationCount++; |
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numb=numb1; |
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while (numb>0) { |
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int digit=numb%10-1; |
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if (digit!=-1&&!uniqueDigits[digit]) { |
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uniqueDigits[digit]=true; |
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} else { |
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goto skip; |
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} |
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carryover(currencyAmts,0); |
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numb/=10; |
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} |
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printf("\n\nCombination count: %d",(combinationCount+2)); |
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numb=numb2; |
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while (numb>0) { |
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int digit=numb%10-1; |
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if (digit!=-1&&!uniqueDigits[digit]) { |
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uniqueDigits[digit]=true; |
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} else { |
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goto skip; |
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} |
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numb/=10; |
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} |
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for (int i=0;i<9;i++) { |
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if (!uniqueDigits[i]) { |
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goto skip; |
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} |
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} |
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//If we got here it's a pandigital.
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if (!uniqueProducts[prod]) { |
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uniqueProducts[prod]=true; |
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sumPandigital+=prod; |
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printf("\n%d*%d = %d is pandigital!",numb1,numb2,prod); |
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} |
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skip: |
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continue; |
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} |
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} |
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printf("\nSum of pandigitals: %ld",sumPandigital); |
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return 0; |
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} |
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