hdu1787——GCD Again(欧拉函数入门)

本文介绍了一道名为“GCDAgain”的ACM竞赛题目，该题要求计算与给定整数N的最大公约数大于1的整数个数。文章详细解释了如何利用欧拉函数进行解答，并提供了一个简洁的C++实现。

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GCD Again

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3583 Accepted Submission(s): 1575

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

题目大意：

给你一个整数N，求范围小于N中的整数中，与N的最大公约数大于1的整数的个数。

思路：

典型的欧拉函数变形。欧拉函数φ(N)是用来求小于N的整数中，与N的最大公约数为1的数的个数。

那么此题的答案ans = N - φ(N) - 1。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
int oula(int n)
{
int res=1,i;
for(i=2; i*i<=n; i++)
if(n%i==0)//i是素数
{
n/=i;//因为要算ni-1次方，所以在这先除一下
res*=i-1;
while(n%i==0)
{
n/=i;//n越来越小
res*=i;
}
}
if(n>1)
res*=n-1;
return res;
}
int main()
{
int n;
while(cin>>n&&n)
{
cout<<n-oula(n)-1<<endl;
}
return 0;
}