1.题目
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
2.代码
#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int euler(int n)
{
int ans=n;
for(int i=2;i<=sqrt(n);i++)
if(n%i==0)
{
ans=ans/i*(i-1);
while(n%i==0)
n/=i;
}
if(n>1) ans=ans/n*(n-1);
return ans;
}
int main()
{
int n;
while(cin>>n)
{
if(n==0) break;
cout<<n-euler(n)-1<<endl;
}
return 0;
}