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FZU 1053 Happy 2004

2009-07-29 20:20 253 查看

Happy 2004

Time Limit:1s	Memory limit:32M
Accepted Submit:142	Total Submit:279

Consider a positive integer X,and let S be the sum of all positive integer divisors of 2004X. Your job is to determine S modulo 29 (the rest of the division of S by 29).

Take X = 1 for an example. The positive integer divisors of 20041 are 1, 2, 3, 4, 6, 12, 167, 334, 501, 668, 1002 and 2004. Therefore S = 4704 and S modulo 29 is equal to 6.

Input

The input consists of several test cases. Each test case contains a line with the integer X (1 <= X <= 10000000).
A test case of X = 0 indicates the end of input, and should not be processed.

Output

For each test case, in a separate line, please output the result of S modulo 29.

Sample Input

1
10000
0

Output for the Sample Input

6
10

Original: 2004 ACM/ICPC Beijing Practice Problem

解题：

运用到欧拉定理，费马小定理。要求某个数A的所有因子和，首先将该数分解：

先贴代码

#include <iostream>
using namespace std;

int happy(int n)
{
	int r,f,t,d;
	f=t=d=1;
	for (r=0;r<=n;++r)
	{
		f=f*(r==n?2:4)%29;		//f表示2^(2n+1)的值
		t=t*3%29;				//t表示3^(n+1)的值
		d=d*22%29;				//d表示167^(n+1)的值
	}
	r=(f-1)*(t-1)*(d-1)%29;		//S=r/332,r的值
	return r*9%29;				//S%29=r*9%29
}
int main()
{
	int n;
	while (cin>>n && n!=0)
	{
		cout<<happy(n%28)<<endl;
	}
	return 0;
}

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