HDOJ 1212 Big Number（同余定理）

Big Number（原题点击）

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7165    Accepted Submission(s): 4945

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
  

 

Author

Ignatius.L

 

Source

杭电ACM省赛集训队选拔赛之热身赛

 同余定理的应用：

（ a + b ) % c = (( a %c ) + ( b % c )) % c ;

（ a * b ) % c = (( a %c ) * ( b % c )) % c ;

参考代码：

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#define MYDD 1103

using namespace std;

int main() {
	char A[MYDD];
	int B,len_A,ans;
	while(scanf("%s %d",A,&B)!=EOF) {
		ans=0;
		len_A=strlen(A);
		for(int i=0; i<len_A; i++) {
			ans=ans*10+(A[i]-'0');
			ans=ans%B;
		}
		printf("%d\n",ans);
	}
	return 0;
}