poj 2407 Relatives 【容斥原理基础题】

Relatives

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11929		Accepted: 5841

Description

Given n, a positive integer, how many positive integers less than n are relatively prime to n? Two integers a and b are relatively prime if there are no integers x > 1, y > 0, z > 0 such that a = xy and b = xz.

Input

There are several test cases. For each test case, standard input contains a line with n <= 1,000,000,000. A line containing 0 follows the last case.

Output

For each test case there should be single line of output answering the question posed above.

Sample Input

7
12
0

Sample Output

6
4

求1到n-1里面与n互质的数的个数。 欧拉打表1000000000太大，用容斥原理。

#include <cstdio>
#include <cstring>
#include <cmath>
#define LL long long
#define MAX 10000000
using namespace std;
int p[1000], k;
void getp(int n)
{
	int i, j;
	k = 0;
	for(i = 2; i*i <= n; i++)
	{
		if(n % i == 0)
		{
			p[k++] = i;
			while(n % i == 0)
			n /= i;
		}
	} 
	if(n > 1) p[k++] = n;
}
int nop(int n)
{
	int top = 0;
	int i, j, que[100000];
	que[top++] = -1;
	for(i = 0; i < k; i++)
	{
		int t = top;
		for(j = 0; j < t; j++)
		que[top++] = que[j]*p[i]*(-1);
	}
	int sum = 0;
	for(i = 1; i < top; i++)
	sum += n/que[i];
	return sum;
}
int main()
{
	int n;
	while(scanf("%d", &n), n)
	{
		getp(n);
		printf("%d\n", n-nop(n));
	}
	return 0;
}