HDOJ-----1787欧拉函数

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

输入一个数n，设 0 < m < n，n和m最大公约数大于1，输出一共有多少组m，

欧拉函数~~

ps:欧拉函数简单性质，一个数x的欧拉值>= y，则这个数最小值为y+1往后的第一个素数，

例：euler(x) >= 8，则x最小为11

#include<cstdio>
int gcd(int a)
{
int ans = a, i;
for(i = 2; i * i <= a; i++)
{
if(a % i == 0)
{
ans -= ans / i;
while(a % i == 0)
{
a /= i;
}
}
}
return a > 1 ? ans - ans / a : ans;//用欧拉函数求出所有互质的数，即所有最大公约数为1的数，再用总数减去
}
int main()
{
int n;
while(scanf("%d", &n) == 1 && n)
{
printf("%d\n", n - gcd(n) - 1);
}
}