poj 2407 Relatives 【容斥原理基础题】

Relatives

Time Limit: 1000MS Memory Limit: 65536KTotal 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

7120

Sample Output

64

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 10000000using 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;}