hdu 1212 Big Number

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3386    Accepted Submission(s): 2305


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
 
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstdlib>
 4 #include <cmath>
 5 #include <cstring>
 6 using namespace std;
 7 char a[1000+10]; int b;
 8 int main(void){
 9 #ifndef ONLINE_JUDGE
10   freopen("1212.in", "r", stdin);
11 #endif
12   while (~scanf("%s%d", a, &b)){
13     int len = strlen(a); int x = a[len-1]-'0', po = 1;
14     x = x % b; int y;
15     for (int i = len - 2; i >= 0; --i){
16       y = a[i] - '0'; po = (po * 10) % b;
17       x = (x + y*po) % b;
18     }
19     printf("%d\n", x);
20   }
21   return 0;
22 }

看了讨论版学会的,不过还是有一点儿小问题,学一下数论吧……

posted on 2013-03-12 22:01  aries__liu  阅读(207)  评论(0编辑  收藏  举报