Solution to problem
number 22 of Project Euler.
Question # 22
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.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.What is the total of all the name scores in the file?
Solution # 22
/***********************************************************************************************************/
#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
#include<time.h>
#include<string.h>
void sort(char**,int);
int main()
{
FILE
*f;
char
c,**str;
double
sum=0;
long
sum1=0;
int
counter=1,i,j,max=0,flag=0; // counter contains the total number of names in
the file
f=fopen("D:/f.txt","r");
//here f.txt contains all the names given in names.txt
if(!f)
printf("\nTHE
FILE OPERATION WAS NOT SUCCESSFUL !\n");
else
{
while(!feof(f))
{
fscanf(f,"
%c",&c);
if(c==',')
{
if(flag>max)
max=flag;
flag=0;
counter++;
}
else
flag++;
}
max-=2;
str=(char**)malloc(sizeof(char*)*counter);
for(i=0;i<counter;i++)
str[i]=(char*)malloc(sizeof(char)*max);
//
counter contains the total number of names in the file
//
max contains number of characters in the longest name
i=0;j=0;
rewind(f);
while(!feof(f))
{
fscanf(f,"
%c",&c);
if(c==',')
{
str[i][j]='\0';
i++;j=0;
continue;
}
if(c=='\"')
continue;
str[i][j++]=c;
}
str[i][j]='\0';
sort(str,counter);
sum=0;
for(i=0,j=0;i<counter;i++,j=0)
{
sum1=0;
while(str[i][j])
{
sum1=sum1+(int)str[i][j]-64;
j++;
}
sum+=(i+1)*sum1;
}
printf("Answer
= %f",sum);
}
printf("\nEXECUTION
TIME = %f\n",clock()/(float)CLK_TCK);
system("pause");
}
void sort(char**str,int counter)
{
int
i,j;
char*tmp;
for(i=1;i<counter;i++)
{
for(j=0;j<counter-i;j++)
{
if(strcmp(str[j],str[j+1])>0)
{
tmp=str[j];
str[j]=str[j+1];
str[j+1]=tmp;
}
}
}
}
/***********************************************************************************************************/
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