【HDU】3501 - Calculation 2（欧拉函数，互质数之和公式）

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Calculation 2

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3564 Accepted Submission(s): 1484

Problem Description

Given a positive integer N, your task is to calculate the sum of the positive integers less than N which are not coprime to N. A is said to be coprime to B if A, B share no common positive divisors except 1.

Input

For each test case, there is a line containing a positive integer N(1 ≤ N ≤ 1000000000). A line containing a single 0 follows the last test case.

Output

For each test case, you should print the sum module 1000000007 in a line.

Sample Input

   3
4
0

Sample Output

   0
2

Author

GTmac

Source

2010 ACM-ICPC Multi-University Training Contest（7）——Host by HIT

如果暴力用容斥原理计算会超时，但是我觉得思路不错。

然后说一下这道题的正解：在数论书上有个公式，计算小于 n 且与 n 互质的数的和。

公式如下：n * Eular ( n ) / 2 （Eular（）表示 n 的欧拉函数值）

代码如下：

#include <cstdio>
__int64 mod = 1e9+7;
int Eular(int n)
{
	int ans = n;
	for (int i = 2 ; i * i <= n ; i++)
	{
		if (n % i == 0)
		{
			ans -= ans / i;
			while (n % i == 0)
				n /= i;
		}
	}
	if (n > 1)
		ans -= ans / n;
	return ans;
}
int main()
{
	int n;
	__int64 ans;
	while (~scanf ("%d",&n) && n)
	{
		if (n == 1)
		{
			printf ("0\n");
			continue;
		}
		ans = ((__int64)(n - 1) * n / 2 - (__int64)n * Eular(n) / 2) % mod;		//套用公式 
		printf ("%I64d\n",ans);
	}
	return 0;
}