HDU1212 Big Number【大数+模除】

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8835    Accepted Submission(s): 5994

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

 

Author

Ignatius.L

 

Source

杭电ACM省赛集训队选拔赛之热身赛

问题链接：HDU1212 Big Number

问题简述：（略）

问题分析：这是一个大数模除问题，需要知道的的公式是(a + b) % m = (a % m + b % m) % m和a * b %m = (a % m * b % m) % m。所以，在从数字字符拼装成大数的过程中就可以不断地进行模除，使得大数模除运算变成小数运算。

程序说明：（略）

题记：（略）

参考链接：（略）

AC的C语言程序如下：

/* HDU1212 Big Number */#include <stdio.h>#define BASE 10#define N 1000char a[N + 1];int main(void){    int b, ans, i;    while(~scanf("%s%d", a, &b)) {        ans = 0;        i = 0;        while(a[i]) {            ans *= BASE;            ans %= b;            ans += a[i] - '0';            i++;        }        printf("%d\n", ans % b);    }    return 0;}