HDU 1787 GCD Again （欧拉函数）

GCD Again

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2920    Accepted Submission(s): 1260


[align=left]Problem Description[/align]
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!

 

[align=left]Input[/align]
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.

 

[align=left]Output[/align]
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.

 

[align=left]Sample Input[/align]

2
4
0

 

[align=left]Sample Output[/align]

0
1

 

[align=left]Author[/align]
lcy
 

[align=left]Source[/align]
2007省赛集训队练习赛（10）_以此感谢DOOMIII

 

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 题意：求小于N的与N互质的数的个数
思路：欧拉函数的应用。

#include<stdio.h>
int euler(int n)
{
int res=n,i;
for(i=2;i*i<=n;i++)
{
if(n%i==0)
res=res/i*(i-1);
while(n%i==0)
n=n/i;//保证n一定是素数
}
if(n>1)
{
res=res/n*(n-1);
}
return res;
}
int main()
{
int n;
while(scanf("%d",&n),n)
{
printf("%d\n",n-euler(n)-1);
}
return 0;
}