HDU 1212 Big Number-优快云博客

本文链接：https://blog.youkuaiyun.com/LuRiCheng/article/details/52556756

本文介绍了一种简单的大数取小数模的算法实现，通过逐位读取字符串形式的大数并进行取模运算，有效地解决了大数运算中的取模问题。该算法适用于ACM竞赛等场景。

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Big Number
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7442    Accepted Submission(s): 5149


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

大数%小数的水题

#include<iostream>
#include<stdlib.h>
#include<stdio.h>
#include<string>
#include<vector>
#include<deque>
#include<queue>
#include<algorithm>
#include<set>
#include<map>
#include<stack>
#include<time.h>
#include<math.h>
#include<list>
#include<cstring>
#include<fstream>
#include<bitset>
//#include<memory.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define pii pair<int,int>
#define INF 1000000007

int main()
{
    //freopen("/home/lu/文档/r.txt","r",stdin);
    //freopen("/home/lu/文档/w.txt","w",stdout);
    string a;
    int b;
    while(cin>>a>>b){
        int ans=0;
        for(int i=0;i<a.size();++i)
            ans=(ans*10+a[i]-'0')%b;
        cout<<ans<<endl;
    }
    return 0;
}