diff --git a/archives/23/current b/archives/23/current new file mode 100755 index 0000000..edfb57c Binary files /dev/null and b/archives/23/current differ diff --git a/archives/23/src/main.c b/archives/23/src/main.c new file mode 100644 index 0000000..beb7bf9 --- /dev/null +++ b/archives/23/src/main.c @@ -0,0 +1,63 @@ +#include +#include "utils.h" +#include + +/* + A perfect number is a number for which the sum of its proper divisors is exactly equal to the number. For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number. + + A number n is called deficient if the sum of its proper divisors is less than n and it is called abundant if this sum exceeds n. + + As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest number that can be written as the sum of two abundant numbers is 24. By mathematical analysis, it can be shown that all integers greater than 28123 can be written as the sum of two abundant numbers. However, this upper limit cannot be reduced any further by analysis even though it is known that the greatest number that cannot be expressed as the sum of two abundant numbers is less than this limit. + + Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers. + + https://projecteuler.net/problem=23 +*/ + +int main(int argc,char**argv) { + //First find all abundant numbers. + //Then iterate through all abundant numbers and determine which numbers are made with them. Numbers not found are to be summed together. + int*abundantNumbers=malloc(sizeof(int)*0); + int abundantNumberCount=0; + for (int i=1;i<=28123;i++) { + int factorSum=1; + //Iterate through divisors. + int max=i/2; + for (int j=2;ji) { + //printf("%d is abundant...\n",i); + abundantNumbers=realloc(abundantNumbers,sizeof(int)*(++abundantNumberCount)); + abundantNumbers[abundantNumberCount-1]=i; + } + } + //Next is to sum up all integers up to 28124. + long sum=0; + for (int i=0;i<28124;i++) { + sum+=i; + } + printf("Sum of all numbers is %ld\n",sum); + + boolean removed[28124] = {}; + + for (int i=0;i +#include + +struct String mult(struct String numb1, struct String numb2) { + struct String n1 = numb1; + struct String n2 = numb2; + byte carryover = 0; + if (numb2.length>numb1.length) { + n1=numb2; + n2=numb1; + } + int addends[n2.length][n1.length+1]; + for (int i=0;i=0;i--) { + carryover=0; + for (int j=n1.length-1;j>=0;j--) { + int mult = (n1.str[j]-'0')*(n2.str[i]-'0')+((carryover!=0)?carryover:0); + //printf(" %d/%d\n",mult,carryover); + carryover=0; + if (mult>=10) { + carryover=mult/10; + mult=mult%10; + } + addends[(n2.length-1)-i][j+1]=mult; + } + if (carryover>0) { + addends[(n2.length-1)-i][0]=carryover; + } + } + //printIntDoubleArr(n2.length,n1.length+1,addends); + struct String sum = {1,"0"}; + for (int i=0;i=numb2.length) { + for (int offset=0;offsetoffset) { + //printf("%c %c\n",numb1.str[numb1.length-offset-1],numb2.str[numb2.length-offset-1]); + int sum=((numb1.str[numb1.length-offset-1]-'0')+(numb2.str[numb2.length-offset-1]-'0'))+((carryover>0)?carryover--:0); + if (sum>=10) { + carryover=1; + sum-=10; + } + str[offset]=sum+'0'; + } else { + str[offset]=numb1.str[numb1.length-offset-1]+((carryover>0)?carryover--:0); + } + //str[offset+1]='\0'; + } + } else { + for (int offset=0;offsetoffset) { + //printf("%c %c\n",numb1.str[numb1.length-offset-1],numb2.str[numb2.length-offset-1]); + int sum=((numb1.str[numb1.length-offset-1]-'0')+(numb2.str[numb2.length-offset-1]-'0'))+((carryover>0)?carryover--:0); + if (sum>=10) { + carryover=1; + sum-=10; + } + str[offset]=sum+'0'; + } else { + str[offset]=numb2.str[numb2.length-offset-1]+((carryover>0)?carryover--:0); + } + //str[offset+1]='\0'; + } + } + if (carryover>0) { + str = realloc(str,++digitCount); + str[digitCount-1]='1'; + //str[digitCount]='\0'; + } + for (int i=0;i #include "utils.h" #include -#include /* - Using names.txt (right click and 'Save Link/Target As...'), a 46K text file containing over five-thousand first names, begin by sorting it into alphabetical order. Then working out the alphabetical value for each name, multiply this value by its alphabetical position in the list to obtain a name score. + A perfect number is a number for which the sum of its proper divisors is exactly equal to the number. For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number. - For example, when the list is sorted into alphabetical order, COLIN, which is worth 3 + 15 + 12 + 9 + 14 = 53, is the 938th name in the list. So, COLIN would obtain a score of 938 × 53 = 49714. + A number n is called deficient if the sum of its proper divisors is less than n and it is called abundant if this sum exceeds n. - What is the total of all the name scores in the file? + As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest number that can be written as the sum of two abundant numbers is 24. By mathematical analysis, it can be shown that all integers greater than 28123 can be written as the sum of two abundant numbers. However, this upper limit cannot be reduced any further by analysis even though it is known that the greatest number that cannot be expressed as the sum of two abundant numbers is less than this limit. - https://projecteuler.net/problem=22 -*/ + Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers. -int nameScore(int pos,char*str) { - int myScore=0; - int marker=0; - while (str[marker]!='\0') { - myScore+=str[marker]-'A'+1; - marker++; - } - return myScore*pos; -} + https://projecteuler.net/problem=23 +*/ int main(int argc,char**argv) { - FILE*file = fopen("p022_names.txt","r"); - int pointer=0; - int nameCount=0; - while (!feof(file)) { - if (fgetc(file)=='"') { - while (fgetc(file)!='"') { + //First find all abundant numbers. + //Then iterate through all abundant numbers and determine which numbers are made with them. Numbers not found are to be summed together. + int*abundantNumbers=malloc(sizeof(int)*0); + int abundantNumberCount=0; + for (int i=1;i<=28123;i++) { + int factorSum=1; + //Iterate through divisors. + int max=i/2; + for (int j=2;ji) { + //printf("%d is abundant...\n",i); + abundantNumbers=realloc(abundantNumbers,sizeof(int)*(++abundantNumberCount)); + abundantNumbers[abundantNumberCount-1]=i; } } - fseek(file,0,0); - char*names[nameCount]; - int currentCount=0; - while (!feof(file)) { - if (fgetc(file)=='"') { - //Start reading the name. - char*newName=malloc(1); - char c=' '; - int length=1; - while ((c=fgetc(file))!='"') { - newName=realloc(newName,++length); - newName[length-2]=c; - newName[length-1]='\0'; - } - if (currentCount<2) { - if (currentCount==1) { - if (strcmp(names[0],newName)<0) { - names[1]=newName; - } else { - names[1]=names[0]; - names[0]=newName; - } - } else { - names[0]=newName; - } - } else { - int low=0; - int high=currentCount; - int pos=0; - int mid=low+(high-low)/2; - while (low<=high) { - mid=low+(high-low)/2; - if (mid>=currentCount) { - break; - } - int diff=strcmp(names[mid],newName); - if (diff<0) { - low=mid+1; - //printf("%s is after %s... New low:%d\n",newName,names[mid],low); - } else { - high=mid-1; - //printf("%s is before %s... New high:%d\n",newName,names[mid],high); - } - } - pos=low+(high-low)/2; - //printf("Position of %s is %d Count:%d.\n",newName,pos,currentCount); - if (pos==currentCount) { - names[currentCount]=newName; - } else { - for (int j=currentCount;j>=pos;j--) { - names[j]=names[j-1]; - } - names[pos]=newName; - } + //Next is to sum up all integers up to 28124. + long sum=0; + for (int i=0;i<28124;i++) { + sum+=i; + } + printf("Sum of all numbers is %ld\n",sum); + + boolean removed[28124] = {}; + + for (int i=0;i