Big Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5961 Accepted Submission(s): 4167
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
Author
Ignatius.L
Source
Recommend
//··A``
1 #include <stdio.h> 2 #include <string.h> 3 char str[1010] ; 4 int main() 5 { 6 int i, n ; 7 while(~scanf("%s %d", str, &n)) 8 { 9 int mod = 0 ; 10 int len = strlen(str) ; 11 for(i = 0; i < len; i++) 12 mod = (mod * 10 + str[i] - '0') % n ; 13 printf("%d ", mod) ; 14 } 15 return 0 ; 16 }