easy n%m

题目描述

I have a very simple problem for you. Given two integers n and m, your job is to calculate the n % m.

输入

The input will consist of a series of pairs of integers n and m, separated by a space, one pair of integers per line. n<=10^100,1<=m<=109

输出

For each pair of input integers a and b you should output the n % m in one line, and with one line of output for each line in input.

样例输入

98757550230217155706156731058600777176 867
837742186949589961446822476827333 15260
751432949647353304314590769697032412519700 16334

样例输出

366
10753
7150

#include<iostream>
#include<string>
const int maxn=10000;
using namespace std;
int main()
{ios::sync_with_stdio(false);//小妹的习惯不用管 cin.tie(0);cout.tie(0);string a;//用字符串来存储大数 int mod;//这是你要取余的数 while(cin>>a>>mod){int l=a.size();int s[maxn];int ans=0;for(int i=0;i<l;i++)//这个就是把字符型数字变成整型数字 {s[i]=a[i]-'0';}for(int i=0;i<l;i++)//每步一取模 {ans=(ans*10+s[i])%mod;}cout<<ans<<endl;}return 0;
}