GCD Again
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1542 Accepted Submission(s): 578
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
Author
lcy
Source
2007省賽集訓隊練習賽（10）_以此感謝DOOMIII
Recommend
lcy
題意：
求小於n的gcd(i,n)大於1的個數
思路 ： 歐拉函數直接求gcd(i,n)==1的個數 用n減即可
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#include<stdio.h>
#include<math.h>
__int64 euler(__int64 x)// 就是公式
{
__int64 i, res=x;
for (i = 2; i <(__int64)sqrt(x * 1.0) + 1; i++)
if(x%i==0)
{
res = res /(__int64)i*(i - 1);
while (x % i == 0) x /= i; // 保證i一定是素數
}
if (x > 1) res = res / (__int64)x*(x - 1);//這裡小心別溢出了
return res;
}
int main()
{
__int64 n;
while(scanf("%I64d",&n)!=EOF)
{
if(!n) break;
printf("%I64d\n",n-euler(n)-1);
}
return 0;
}
