Big Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2407 Accepted Submission(s): 1639
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
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Eddy
Statistic | Submit | Discuss | Note
解题报告：这是一道数论题！就是求AmodB，但是A是大数，想了好半天，也没有想出神马思路，就借鉴了一下网上“大牛”们的思路
下面就是我在网上找的简单方法：
用递推的思想，就可以总结出下面的公式；
举例：
12345 9
余数等于（12340%9+5%9）%9；
而12340 9
（12300%9+40%9）%9;
依次...
最后（10000%9+2000%9）%9；
而10000%9=（1%9*10000）%9
2000%9=（2%9*1000）%9
即（1*10+2）%9*1000%9；
即可得到：for（i=0;i<len;i++）
{
sum=sum*10+s[i]-'0';
sum=sum%9;
}
我的代码如下：
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N = 1010;
char str[N];
int main()
{
int ans,i,len,m;
while(scanf("%s%d",str,&m)!=EOF)
{
len=strlen(str);
ans=0;
for(i=0;i<len;i++)
{
ans=ans*10+str[i]-'0';
ans=ans%m;
}
printf("%d\n",ans);
}
return 0;
}