欧拉函数的简单应用

最新推荐文章于 2019-05-07 12:18:00 发布

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#Given n a positive

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本文探讨了如何计算小于给定正整数n的所有与n互质的正整数的数量。互质定义为两个整数a和b不存在大于1的整数x使得a和b都能被x整除。文章通过示例输入输出展示了程序处理此类问题的方法。

Given n, a positive integer, how many positive integers less than n are relatively prime to n? Two integers a and b are relatively prime if there are no integers x > 1, y > 0, z > 0 such that a = xy and b = xz.

Input

There are several test cases. For each test case, standard input contains a line with n <= 1,000,000,000. A line containing 0 follows the last case.

Output

For each test case there should be single line of output answering the question posed above.

Sample Input

7
120

Sample Output

6
4

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;  
int a[1000010];   
  int main()  
{  
	int t;
    int num=0,sum=0,max=0;  
    cin >> t;  
     
        while(t--)  
        {  
        	int n,j=0;  
            cin >> n;  
            for(int i=0;i<n;i++)  
            {  
                cin  >>  a[i] ;  
                if(max<a[i]) 
					max=a[i];  
            }   
            for(j;j<n;j++)  
            {  
                if(a[j]!=max) 
					sum+=a[j];  
                if(sum>=max-1)
				 	break;  
            }  
            if(j==n) 
            	cout << "No" <<endl;
            else 
            	cout << "Yes" <<endl;  
        }  
    
    return 0;  
}