hdu 1787 GCD Again 欧拉函数小水水数论

最新推荐文章于 2021-08-04 15:15:41 发布

chasexie(xiehonghao)

最新推荐文章于 2021-08-04 15:15:41 发布

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本文介绍了一个ACM竞赛问题GCDAgain，目标是计算小于给定整数N的所有整数M中，满足M与N的最大公约数大于1的数量。文章提供了一种基于欧拉函数的方法来解决该问题，并给出了具体的实现代码。

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GCD Again

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1542 Accepted Submission(s): 578

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

Sample Output

0
1

Author

lcy

Source

2007省赛集训队练习赛（10）_以此感谢DOOMIII

Recommend

lcy

题意：

求小于n的gcd(i,n)大于1的个数

思路：欧拉函数直接求gcd(i,n)==1的个数用n减即可

#include<stdio.h>
#include<math.h>
__int64 euler(__int64 x)// 就是公式   
{  
    __int64 i, res=x;  
    for (i = 2; i <(__int64)sqrt(x * 1.0) + 1; i++)  
        if(x%i==0)  
        {  
            res = res /(__int64)i*(i - 1);
			
            while (x % i == 0) x /= i; // 保证i一定是素数    
        }  
        if (x > 1) res = res / (__int64)x*(x - 1);//这里小心别溢出了   
        return res;  
}  

int main()
{
	__int64 n;
    while(scanf("%I64d",&n)!=EOF)
	{
		if(!n) break;
		printf("%I64d\n",n-euler(n)-1);
	}
	 return 0;
}