Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5988    Accepted Submission(s): 4184

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

2 312 7152455856554521 3250

 

Sample Output

251521
C语言程序代码
/*解题思路：：这是个大数题，要用数组。有数学上的一些规律即：（a+b）%c=(a%c+b%c)%c;a*b%c=(a%c*b%c)%c;所以根据这个就可以将大数变小， 用数组逐位取余即可，避免计算大数。 */#include<stdio.h>#include<string.h>#include<math.h>int main(){char s[100001];int a,n,m,sum,l,i;while(scanf("%s %d",s,&n)!=EOF){l=strlen(s);sum=0;for(i=0;i<l;i++)sum=(sum*10+(s[i]-'0')%n)%n;printf("%d\n",sum);}return 0;}