(hdu step 7.2.2)GCD Again(欧拉函数的简单应用——求[1,n)中与n不互质的元素的个数)

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本文介绍了一道名为“GCDAgain”的ACM竞赛题目，该题要求计算小于N的所有正整数M中与N的最大公约数大于1的数量。文章详细解释了如何利用欧拉函数求解，并提供了具体的实现代码。

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题目：

GCD Again

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 125 Accepted Submission(s): 84

Problem Description

Do you have spent some time to think and try to solve those unsolved problem after one ACM contest?
No? Oh, you must do this when you want to become a "Big Cattle".
Now you will find that this problem is so familiar:
The greatest common divisor GCD (a, b) of two positive integers a and b, sometimes written (a, b), is the largest divisor common to a and b. For example, (1, 2) =1, (12, 18) =6. (a, b) can be easily found by the Euclidean algorithm. Now I am considering a little more difficult problem:
Given an integer N, please count the number of the integers M (0<M<N) which satisfies (N,M)>1.
This is a simple version of problem “GCD” which you have done in a contest recently,so I name this problem “GCD Again”.If you cannot solve it still,please take a good think about your method of study.
Good Luck!

Input

Input contains multiple test cases. Each test case contains an integers N (1<N<100000000). A test case containing 0 terminates the input and this test case is not to be processed.

Output

For each integers N you should output the number of integers M in one line, and with one line of output for each line in input.

Sample Input

2
4
0

Sample Output

0
1

Author

lcy

Source

2007省赛集训队练习赛（10）_以此感谢DOOMIII

Recommend

lcy

题目分析：

欧拉函数的简单应用。本体先使用phi(n)求出[1,n]中与n互质的元素的个数,然后再使用n-phi(n)求出[1,n]中与

n不互质的元素的个数即可，最后还需要把它自己给减掉。也就是n-phi(n)-1.

这道题需要的需要注意的是：

1、在这里,我们还回顾一下"互质"的定义：

互质，公约数只有1的两个整数，叫做互质整数·公约数只有1的两个自然数，叫做互质自然数，后者是前者的特殊情形·。

2、关于使用预处理的方式来求欧拉值和使用phi(n)来求欧拉值得两种方式的选择的个人考虑：

1）当n比较小，同一个输入样例需要多次用到phi[i]时,这时可以考虑使用预处理的方式。如果当n比较大的时候仍使用这种方式,很可能会直接MLE，如这道题。

2）当n比较大,同一个输入样例只需要使用一个phi[i]时,这是我们可以考虑使用调用phi(i)的方式。

代码如下：

#include <iostream>
#include <stdio.h>
#include <string>
#include <cmath>
#include <algorithm>
using namespace std;

typedef unsigned long long int longint;

longint phi(longint num) {
	longint sum = 1;
	for (long int i = 2; i <= sqrt((double long) num); i++) {
		if (num % i == 0) {
			while (num % i == 0) {
				sum *= i;
				num /= i;
			}
			sum /= i;
			sum *= (i - 1);
		}
	}

	if (num != 1) {
		sum *= (num - 1);
	}

	return sum;
}

int main(){
	int n;
	while(scanf("%d",&n)!=EOF,n){
		/**
		 * 最后为什么要减1呢?
		 * 因为这道题要求的是[1,n)中与n不互质的元素的个数,
		 * 需要把n自己给减掉.
		 */
		printf("%lld\n",n - phi(n) - 1);
	}

	return 0;
}