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.
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.
For each test case, you have to ouput the result of A mod B.
2 3
12 7
152455856554521 3250
#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!!!!!!