HDU 1212 Big Number

             HDU 1212 Big Number

          Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
               Total Submission(s): 9158    Accepted Submission(s): 6214


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
 
思路:大数求余:
        ( a + b ) % c = ( ( a % c) + ( b % c ) ) % c;
        ( a * b ) % c = ( ( a % c ) * ( b % c ) ) % c;
 
(来自 百度文库 https://wenku.baidu.com/view/3784592beefdc8d376ee32ba.html )
#include<iostream>
#include<string>
using namespace std;
int main() {
    int n, i, j;
    string s;
    while(cin>>s>>n) {
        j = 0;
        for(i = 0; i < s.length(); i++) {
            j = ( j*10 % n + (s[i]-'0')%n ) % n;
        }
        cout<<j<<endl;
    }
    return 0;
}

 

posted @ 2018-04-03 16:45  落云小师妹  阅读(145)  评论(5编辑  收藏  举报