大数取余

发布时间：2020-12-14 03:06:39 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): 4680????Accepted Submission(s): 3247 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): 4680????Accepted Submission(s): 3247

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
  
  
  
  
   
   ?
  
  
  
  
   
   #include<stdio.h>
#include<string.h>
int mod(char *n1,int n2)
{
? int temp,i;
? int k=strlen(n1);
????? temp=0;
? for(i=0;i<k;i++)
? {
??? temp=temp*10+(n1[i]-'0');
??? temp%=n2;
? }??? 
? return temp;
}
int main()
{
??? int n,;
??? char str[10000];
??? while(~scanf("%s %d*c",str,&n))
??? {
?????? printf("%dn",mod(str,n));???????????????????????????? 
??? }
return 0;??? 
}

（编辑：李大同）

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