HDU 1787 欧拉公式

GCD Again

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1801    Accepted Submission(s): 686

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

 

欧拉公式：f(n) = n * (1-1/p1) * (1-1/p2)...* (1-1/pk)，其中p1到pk是n的所有不同质因数，比如12 = 2*2*3；此时只取2和3，也就是f(12)=12*(1-1/2)*(1-1/3) = 4;f(n)的值就是比n小的且与n互质的数的个数，f(12)=4也就是在12以内里，有4个与12互质的数（1，5，7，11）。

思路：根据题目要求，求出f(n),用n减去互质的数f(n)，再减去它自己本身，即为答案。

View Code

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6  7 using namespace std; 8  9 int euler(int n)10 {11     int ret = n;12     int t = (int)sqrt(n*1.0);13     for(int i=2; i<=t; i++)14     {15         if(n%i == 0)16         {17             ret = ret/i * (i-1);18             while(n%i == 0)19                 n /=  i;20         }21     }22     if(n > 1)23         ret = ret/n * (n-1);24     return ret;25 }26 27 int main()28 {29     int N;30     while(~scanf("%d",&N) && N)31     {32         printf("%d\n",N-1-euler(N));33     }34     return 0;35 }