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046318072f
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#include <stdio.h> |
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#include "utils.h" |
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#define TARGET 2 |
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/*
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By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23. |
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3 |
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7 4 |
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2 4 6 |
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8 5 9 3 |
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That is, 3 + 7 + 4 + 9 = 23. |
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Find the maximum total from top to bottom of the triangle below: |
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75 |
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95 64 |
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17 47 82 |
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18 35 87 10 |
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20 04 82 47 65 |
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19 01 23 75 03 34 |
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88 02 77 73 07 63 67 |
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99 65 04 28 06 16 70 92 |
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41 41 26 56 83 40 80 70 33 |
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41 48 72 33 47 32 37 16 94 29 |
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53 71 44 65 25 43 91 52 97 51 14 |
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70 11 33 28 77 73 17 78 39 68 17 57 |
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91 71 52 38 17 14 91 43 58 50 27 29 48 |
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63 66 04 68 89 53 67 30 73 16 69 87 40 31 |
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04 62 98 27 23 09 70 98 73 93 38 53 60 04 23 |
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NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o) |
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https://projecteuler.net/problem=18
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*/ |
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int main(int argc,char**argv) { |
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int rows[][15] = { |
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{75} |
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,{95,64} |
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,{17,47,82} |
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,{18,35,87,10} |
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,{20,4,82,47,65} |
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,{19,1,23,75,3,34} |
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,{88,2,77,73,7,63,67} |
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,{99,65,4,28,6,16,70,92} |
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,{41,41,26,56,83,40,80,70,33} |
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,{41,48,72,33,47,32,37,16,94,29} |
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,{53,71,44,65,25,43,91,52,97,51,14} |
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,{70,11,33,28,77,73,17,78,39,68,17,57} |
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,{91,71,52,38,17,14,91,43,58,50,27,29,48} |
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,{63,66,4,68,89,53,67,30,73,16,69,87,40,31} |
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,{4,62,98,27,23,9,70,98,73,93,38,53,60,4,23}}; |
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int max=14; |
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for (int row=14;row>0;row--) { |
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for (int col=0;col<max;col++) { |
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int numb1=rows[row][col]; |
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int numb2=rows[row][col+1]; |
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//printf("Picked numbers: %d %d\n",numb1,numb2);
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rows[row-1][col]+=(numb1>numb2)?numb1:numb2; |
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} |
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max--; |
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} |
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for (int i=0;i<15;i++) { |
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for (int j=0;j<15;j++) { |
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printf("%d\t",rows[i][j]); |
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} |
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printf("\n"); |
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} |
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printf("\nYour maximum route is %d",rows[0][0]); |
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return 0; |
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} |
@ -0,0 +1,54 @@ |
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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 add(struct String numb1, struct String numb2){ |
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byte carryover=0; |
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int digitCount=0; |
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char*str = malloc(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]; |
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} |
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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]; |
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} |
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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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} |
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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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struct String newStr = {digitCount,str}; |
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return newStr; |
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} |
@ -0,0 +1,9 @@ |
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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); |
@ -1,168 +1,76 @@ |
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#include <stdio.h> |
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#include "utils.h" |
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#define TARGET 2 |
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/*
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If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total. |
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By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23. |
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3 |
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7 4 |
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2 4 6 |
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8 5 9 3 |
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If all the numbers from 1 to 1000 (one thousand) inclusive were written out in words, how many letters would be used? |
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That is, 3 + 7 + 4 + 9 = 23. |
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Find the maximum total from top to bottom of the triangle below: |
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NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 115 (one hundred and fifteen) contains 20 letters. The use of "and" when writing out numbers is in compliance with British usage. |
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75 |
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95 64 |
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17 47 82 |
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18 35 87 10 |
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20 04 82 47 65 |
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19 01 23 75 03 34 |
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88 02 77 73 07 63 67 |
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99 65 04 28 06 16 70 92 |
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41 41 26 56 83 40 80 70 33 |
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41 48 72 33 47 32 37 16 94 29 |
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53 71 44 65 25 43 91 52 97 51 14 |
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70 11 33 28 77 73 17 78 39 68 17 57 |
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91 71 52 38 17 14 91 43 58 50 27 29 48 |
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63 66 04 68 89 53 67 30 73 16 69 87 40 31 |
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04 62 98 27 23 09 70 98 73 93 38 53 60 04 23 |
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https://projecteuler.net/problem=17
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NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o) |
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https://projecteuler.net/problem=18
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*/ |
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int main(int argc,char**argv) { |
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int totalCount=0; |
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int counter=1; |
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int i=1; |
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int hundredsDigit=-1; |
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int tensDigit=-1; |
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while (i<=1000) { |
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hundredsDigit=-1; |
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tensDigit=-1; |
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counter=i; |
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int currentCount=totalCount; |
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if (counter==1000) { |
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//onethousand (11)
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totalCount+=11; |
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counter-=1000; |
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printf("onethousand"); |
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} |
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if (counter>=100) { |
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hundredsDigit=counter/100; |
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switch (hundredsDigit) { |
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case 1: |
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case 2: |
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case 6:{ |
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//onehundred (10)
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totalCount+=10; |
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printf("%s",hundredsDigit==1?"one":hundredsDigit==2?"two":"six"); |
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}break; |
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case 3: |
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case 7: |
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case 8:{ |
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//threehundred (12)
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totalCount+=12; |
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printf("%s",hundredsDigit==3?"three":hundredsDigit==7?"seven":"eight"); |
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}break; |
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case 4: |
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case 5: |
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case 9:{ |
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//fourhundred (11)
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totalCount+=11; |
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printf("%s",hundredsDigit==4?"four":hundredsDigit==5?"five":"nine"); |
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}break; |
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} |
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printf("hundred"); |
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counter-=hundredsDigit*100; |
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} |
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if (counter>=10) { |
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tensDigit=counter/10; |
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if (tensDigit!=0) { |
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if (hundredsDigit!=-1) { |
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//and (3)
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totalCount+=3; |
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printf("and"); |
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} |
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if (counter<20) { |
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switch (counter) { |
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case 10:{ |
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//ten (3)
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totalCount+=3; |
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printf("ten"); |
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}break; |
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case 11:{ |
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//eleven (6)
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totalCount+=6; |
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printf("eleven"); |
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}break; |
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case 12:{ |
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//twelve (5)
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totalCount+=5; |
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printf("twelve"); |
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}break; |
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case 13: |
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case 14: |
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case 19:{ |
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//thirteen (8)
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totalCount+=8; |
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printf("%s",counter==13?"thirteen":counter==14?"fourteen":"nineteen"); |
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}break; |
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case 15: |
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case 16:{ |
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//fifteen (7)
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totalCount+=7; |
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printf("%s",counter==15?"fifteen":"sixteen"); |
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}break; |
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case 17: |
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case 18:{ |
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//seventeen (9)
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totalCount+=9; |
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printf("%s",counter==17?"seventeen":"eighteen"); |
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}break; |
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} |
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counter=0; |
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} else { |
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switch (tensDigit) { |
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case 2: |
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case 3: |
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case 8: |
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case 9:{ |
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//twenty (6)
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totalCount+=6; |
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printf("%s",tensDigit==2?"twenty":tensDigit==3?"thirty":tensDigit==8?"eighty":"ninety"); |
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}break; |
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case 4: |
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case 5: |
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case 6:{ |
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//forty (5)
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totalCount+=5; |
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printf("%s",tensDigit==4?"forty":tensDigit==5?"fifty":"sixty"); |
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}break; |
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case 7:{ |
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//seventy (7)
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totalCount+=7; |
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printf("seventy"); |
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}break; |
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} |
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counter-=tensDigit*10; |
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} |
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} |
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} |
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if (counter>0) { |
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if (tensDigit==-1&&hundredsDigit!=-1) { |
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//and (3)
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totalCount+=3; |
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printf("and"); |
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int rows[][15] = { |
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{75} |
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,{95,64} |
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,{17,47,82} |
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,{18,35,87,10} |
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,{20,4,82,47,65} |
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,{19,1,23,75,3,34} |
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,{88,2,77,73,7,63,67} |
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,{99,65,4,28,6,16,70,92} |
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,{41,41,26,56,83,40,80,70,33} |
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,{41,48,72,33,47,32,37,16,94,29} |
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,{53,71,44,65,25,43,91,52,97,51,14} |
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,{70,11,33,28,77,73,17,78,39,68,17,57} |
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,{91,71,52,38,17,14,91,43,58,50,27,29,48} |
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,{63,66,4,68,89,53,67,30,73,16,69,87,40,31} |
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,{4,62,98,27,23,9,70,98,73,93,38,53,60,4,23}}; |
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int max=14; |
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for (int row=14;row>0;row--) { |
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for (int col=0;col<max;col++) { |
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int numb1=rows[row][col]; |
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int numb2=rows[row][col+1]; |
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//printf("Picked numbers: %d %d\n",numb1,numb2);
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rows[row-1][col]+=(numb1>numb2)?numb1:numb2; |
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} |
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switch (counter) { |
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case 1: |
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case 2: |
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case 6:{ |
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//one (3)
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totalCount+=3; |
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printf("%s",counter==1?"one":counter==2?"two":"six"); |
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}break; |
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case 3: |
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case 7: |
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case 8:{ |
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//three (5)
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totalCount+=5; |
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printf("%s",counter==3?"three":counter==7?"seven":"eight"); |
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}break; |
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case 4: |
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case 5: |
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case 9:{ |
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//four (4)
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totalCount+=4; |
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printf("%s",counter==4?"four":counter==5?"five":"nine"); |
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}break; |
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max--; |
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} |
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for (int i=0;i<15;i++) { |
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for (int j=0;j<15;j++) { |
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printf("%d\t",rows[i][j]); |
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} |
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printf(" %d:%d\n",i,totalCount-currentCount); |
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i++; |
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printf("\n"); |
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} |
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printf("%d",totalCount); |
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printf("\nYour maximum route is %d",rows[0][0]); |
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return 0; |
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} |
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