Big Number
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
2 5 1521
Author
Ignatius.L
Source
Recommend
Eddy
在做题之前，先了解这样一些结论：
A*B % C = (A%C * B%C)%C
(A+B)%C = (A%C + B%C)%C
(A+B)%C = (A%C + B%C)%C
如 532 mod 7 =（500%7+30%7+2%7)%7;
当然还有a*b mod c=(a mod c+b mod c)mod c;
如35 mod 3=((5%3)*(7%3))%3
当然还有a*b mod c=(a mod c+b mod c)mod c;
如35 mod 3=((5%3)*(7%3))%3
有了这一些结论，题目就好做了！
代码如下：
#include<iostream>
#include<string.h>
using namespace std;
const int MAX=100010;
int main()
{
char str[MAX];
int s,len,sum;
while(scanf("%s%d",str,&s)!=EOF)
{
len=strlen(str);
sum=0;
for(int i=0;i<len;i++)
sum=(sum*10+(str[i]-'0')%s)%s;
cout<<sum<<endl;
}
return 0;
}