【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

340

 

Sample Output

02

 

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;}