Big Number
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 16 Accepted Submission(s) : 11
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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
杭电ACM省赛集训队选拔赛之热身赛
题目大意：求两个数的余数，因为数比较大所以我们不可以直接用a%b的方法，回想一下中学时我们求大数相除的余数时的方法，先用说的最高位去除除数求余数，然后用余数乘以十加被除数的下一位然后再初除数依次进行下去，直到加到最后一位再求得的余数便是最后结果。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define max 100010
int main()
{
char str[max];
int m,l,sum;
while(scanf("%s%d",str,&m)!=EOF)
{
l=strlen(str);
sum=0;
for(int i=0;i<l;i++)
sum=(sum*10+(str[i]-'0')%m)%m;
cout<<sum<<endl;
}
return 0;
}
