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杭电1212--Big Number

时间:2015-07-24 22:27:24      阅读:185      评论:0      收藏:0      [点我收藏+]

标签:

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.
 

 

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
Eddy   |   We have carefully selected several similar problems for you:  1215 1211 1210 1214 1213 
 
//··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\n", mod) ;
14     }
15     return 0 ;    
16 } 

 

 

杭电1212--Big Number

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原文地址:http://www.cnblogs.com/fengshun/p/4674708.html

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