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.
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
题意:
给出正整数 A 和 B,求 A mod B。
思路:
模拟除法运算即可。
例:
代码:
#include <set> #include <map> #include <list> #include <stack> #include <queue> #include <deque> #include <cmath> #include <string> #include <vector> #include <cstdio> #include <cstring> #include <cstdlib> #include <sstream> #include <iostream> #include <algorithm> #include <unordered_map> #define INF 0x3f3f3f3f #define ll long long #define ull unsigned long long #define fcio ios::sync_with_stdio(false); cin.tie(0); cout.tie(0) using namespace std; int main() { string a; int b; while(cin>>a>>b) { int t=0; for(int i=0;i<a.size();i++) { t=t*10+a[i]-'0'; t=t%b; } cout<<t<<endl; } }