poj 2773 Happy 2006(容斥原理+二分）

Happy 2006

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 9139		Accepted: 3079

DescriptionTwo positive integers are said to be relatively prime to each other if the Great Common Divisor (GCD) is 1. For instance, 1, 3, 5, 7, 9...are all relatively prime to 2006.Now your job is easy: for the given integer m, find the K-th element which is relatively prime to m when these elements are sorted in ascending order.InputThe input contains multiple test cases. For each test case, it contains two integers m (1 <= m <= 1000000), K (1 <= K <= 100000000).OutputOutput the K-th element in a single line.Sample Input

2006 1
2006 2
2006 3

Sample Output

1
3
5

#include <iostream>#include <cstdio>#include <vector>using namespace std;#define ll long longll m , K;vector<int> p;void initial(){p.clear();for(int i = 2; i*i <= m; i++){if(m%i==0){p.push_back(i);while(m%i==0) m /= i;}}if(m > 1) p.push_back(m);}ll Inclusion_Exclusion(ll x){int cnt = p.size();ll ans = 0;for(int i = 1; i < (1<<cnt); i++){ll ret = 1 , sum = 0;for(int j = 0; j < cnt; j++){if(i&(1<<j)){sum++;ret *= p[j];}}if(sum%2==0) ans -= x/ret;else ans += x/ret;}return x-ans;}void computing(){ll l = 0 , r = ((ll)1)<<62;while(l < r){ll mid = (l+r)/2;if(Inclusion_Exclusion(mid) >= K) r = mid;else l = mid+1;}printf("%lld\n" , l);}int main(){while(cin >> m >> K){initial();computing();}return 0;}