朴素求欧拉函数模板(1787)

GCD Again

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2846    Accepted Submission(s): 1217

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

240

 

Sample Output

01

筛法求欧拉函数必须保存在一个数组中，当n很大时，无法保存，所以这里采用朴素的求欧拉函数的方法，但是时间复杂度相对较高了。

/*------------------Header Files------------------*/#include <iostream>#include <cstring>#include <string>#include <cstdio>#include <algorithm>#include <cstdlib>#include <ctype.h>#include <cmath>#include <stack>#include <queue>#include <deque>#include <map>#include <vector>#include <set>#include <limits.h>using namespace std;/*------------------Definitions-------------------*/#define LL long long#define uLL unsigned long long#define PI acos(-1.0)#define INF 0x3F3F3F3F#define MOD 9973#define MAX 1000000005#define lson rt<<1,l,m#define rson rt<<1|1,m+1,r/*---------------------Work-----------------------*/int euler(int n){int ans=n;int m=(int)sqrt(n+0.5);for(int i=2;i<=m;i++){if(n%i==0){ans=ans/i*(i-1);while(n%i==0) n/=i;}}if(n>1) ans=ans/n*(n-1); //这一步不能忘，n可能只有1和n两个因子return ans;}void work(){int n;while(scanf("%d",&n)==1,n){printf("%d\n",n-euler(n)-1);}}/*------------------Main Function------------------*/int main(){//freopen("test.txt","r",stdin);work();return 0;}