HDU1212 Big Num

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7940    Accepted Submission(s): 5465


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   |   We have carefully selected several similar problems for you:   1215  1211  1210  1214  1002 


#include <iostream>
#include <cstring>
using namespace std;
int main()
{
    char a[10002];
    int b;
    while(cin>>a>>b)
    {
        int res=0;
        for(int i=0;i<strlen(a);i++)
            res=(10*res+a[i]-'0')%b;
        cout<<res<<endl;
    }
    return 0;
}

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值