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HDU 1212 Big Number(水题)大数取模

发布时间:2020-12-14 03:29:31 所属栏目:大数据 来源:网络整理
导读:Big Number Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4556????Accepted Submission(s): 3152 Problem Description As we know,Big Number is always troublesome. But it's really impor

Big Number

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4556????Accepted Submission(s): 3152


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

AC代码:

#include<stdio.h>
#include<string.h>
int main(){
	char a[1010];
	int i,len,ans,b;
	while(~scanf("%s%d",a,&b)){
		len=strlen(a);
		ans=0;
		for(i=0;i<len;i++)
			ans=(ans*10+(a[i]-'0'))%b;
		printf("%dn",ans);
	}
	return 0;
}

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