Big Number 求模

Big Number

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 19   Accepted Submission(s) : 14

Problem Description

As we know, Big Number is always troublesome. But it's really important in our ACM. And today, your task is to write a program to calculate A mod B.

To make the problem easier, I promise that B will be smaller than 100000.

Is it too hard? No, I work it out in 10 minutes, and my program contains less than 25 lines.

 

Input

The input contains several test cases. Each test case consists of two positive integers A and B. The length of A will not exceed 1000, and B will be smaller than 100000. Process to the end of file.

 

Output

For each test case, you have to ouput the result of A mod B.

 

Sample Input

2 312 7152455856554521 3250

 

Sample Output

251521

 

  累加求模

用递推的思想，就可以总结出下面的公式；举例：12345 9余数等于（12340%9+5%9）%9；而12340 9（12300%9+40%9）%9;依次...最后（10000%9+2000%9）%9；而10000%9=（1%9*10000）%92000%9=（2%9*1000）%9即（1*10+2）%9*1000%9；即可得到：for（i=0;i<len;i++）           {             sum=sum*10+s[i]-'0';              sum=sum%9;            }#include<stdio.h>#include<string.h>int main(){ int n,sum,i,k; char s[1001]; while(scanf("%s%d",s,&n)!=EOF) {  k=strlen(s);  sum=0;  for(i=0;i<k;i++)  {   sum=sum*10+s[i]-'0';   sum=sum%n;  }  printf("%d\n",sum); } return 0;}