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.
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; }