【hdu 1787】 GCD Again（根号n求phi）

GCD Again

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

Total Submission(s): 3061    Accepted Submission(s): 1348


[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
 

[align=left]Recommend[/align]
lcy   |   We have carefully selected several similar problems for you:  1788 1573 2824 1286 1215 
 
【题解】【根号n求phi】

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int x;
inline int find(int m)
{
int i,k=m,l=m;
for (i=2;i*i<=m;++i)
if (!(l%i))
{
k=k-k/i;
do{
l/=i;
}while(!(l%i));
}
if (l>1) k=k-k/l;
return k;
}
int main()
{
while ((scanf("%d",&x)==1)&&x)
printf("%d\n",x-find(x)-1);
return 0;
}