HDU1212加深下对取模运算的理解

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3771    Accepted Submission(s): 2586

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

 

//原理： (a+b)%c = ((a%c)+(b%c))%c
  //    (a*b)%c = ((a%c)*(b%c))%c
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define maxn 1000+5

int num[250];
char numc[maxn];

int main()
{
    int b;
    while(cin>>numc>>b)
    {
        int i = strlen(numc) - 1;
        int p = 1,ans = 0;
        while(i>=0)
        {
            ans = (ans + (numc[i]-'0')*p)%b;
            p *= 10;
            p %= b;
            i--;
        }
        cout<<ans<<endl;
    }
}