同余定理之杭电1212_Big Number

[align=left]Problem Description[/align]
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.

 

[align=left]Input[/align]
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.

 

[align=left]Output[/align]
For each test case, you have to ouput the result of A mod B.

 

[align=left]Sample Input[/align]

2 3
12 7
152455856554521 3250

 

[align=left]Sample Output[/align]

2
5
1521

 

AC代码如下：

#include "iostream"
using namespace std;

int main(int argc, char* argv[])
{
char a[1001];
int i,b,s,sum;
while(cin>>a>>b)
{
s=strlen(a);
sum=0;
for (i=0;i<s;i++)
{
sum=(sum*10+a[i]-48)%b;
}
cout<<sum<<endl;
}
return 0;
}