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 3
12 7
152455856554521 3250
Sample Output
题意简单，不用讲的呐！
要是你看不懂我非常操心你的四级肿么办！！！！！！
直接上代码，能够作为模板！！！！！！
这代码有点抄袭别人的感觉；
开毛线玩笑，模板啊！！！！！！
模板不一样怎么叫模板！！！！！
#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
int main()
{
int i,j,k;
int t,n,m;
char ch[1006];
while(cin>>ch>>n)
{
m=0;
for(i=0;i<strlen(ch);i++)
{
m*=10;
m+=ch[i]-‘0‘;
m%=n;
}
printf("%d\n",m);
}
return 0;
}
写代码能力有限，如有编程爱好者发现bug，还请指出，不胜感激！！！！！！
