hdu 1212 Big Number(同余定理)
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
解题思路:这是一道大数取模的题目,运用到同余定理:(a+b)%c=(a%c+b%c)%c=(a+b%c)%c。
(a*b)%c=(a%c*b%c)%c。本质是模拟做除法运算,过程中只需保留余数即可!举个例子:572%7=((((5%7==5)*10+7==57)%7==1)*10+2==12)%7==5。
AC代码:
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long LL; 4 string str;int mod, ans; 5 int main() { 6 while(cin >> str >> mod) { 7 ans = 0; 8 for(int i = 0; str[i]; ++i) ans = (ans * 10 + (str[i] - '0')) % mod; 9 cout << ans << endl; 10 } 11 return 0; 12 }