杭电 —1212 大数取模

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7152    Accepted Submission(s): 4934

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

 

Author

Ignatius.L

 

Source

杭电ACM省赛集训队选拔赛之热身赛

 

解题思路：这道题用的求模思路和我们平时进行的除法的套路是一样的。

先是第一个数取模，然后乘10加上后面的数接着取模，往后类推。

AC代码：

#include<stdio.h>#include<string.h>int main(){  char a[1100];  int b,i;  while(scanf("%s%d",a,&b)!=EOF)   {     int len=strlen(a);     int sum=0;  for(i=0;i<len;i++)   {     sum=sum*10+a[i]-'0';     sum=sum%b;   }     printf("%d\n",sum);   }    return 0;}