HDU 1212 Big Number（大数取模）

Big Number

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.

 

[align=left]Input[/align]

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.

 

[align=left]Output[/align]

For each test case, you have to ouput the result of A mod B.

 

[align=left]Sample Input[/align]

2 3
12 7
152455856554521 3250

 

Sample Output

2
5
1521

【思路分析】

   用了大数取模最基本的方法——按位取模。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxn = 100005;
int mod,len,ans;
char s[maxn];
void init()
{
len = strlen(s);
ans = 0;
}
void solve()
{
for(int i = 0;i < len;i++)
{
ans = (ans * 10 + (s[i] - '0') % mod) % mod;
}
printf("%d\n",ans);
}
int main()
{
while(scanf("%s%d",s,&mod) != EOF)
{
init();
solve();
}
return 0;
}