poj 2773 Happy 2006 容斥原理+二分

题目链接

容斥原理求第k个与n互质的数。

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-;
const int mod = 1e9+;
const int inf = ;
const int dir[][] = { {-, }, {, }, {, -}, {, } };
vector<int> v;
void init(int n) {
    v.clear();
    for(int i = ; i*i<=n; i++) {
        if(n%i==) {
            v.pb(i);
        }
        while(n%i==)
            n/=i;
    }
    if(n!=)
        v.pb(n);
}
ll solve(ll x) {
    int len = v.size();
    ll ret = ;
    for(int i = ; i<(<<len); i++) {
        ll cnt = , ans = ;
        for(int j = ; j<len; j++) {
            if((<<j)&i) {
                cnt++;
                ans *= v[j];
            }
        }
        if(cnt&) {
            ret += x/ans;
        } else {
            ret -= x/ans;
        }
    }
    return x-ret;
}
int main()
{
    int n, m;
    while(cin>>n>>m) {
        init(n);
        ll l = , r = 1LL<<, ans;
        while(r>=l) {
            ll mid = (l+r)/;
            ll ret = solve(mid);
            if(ret>=m)
                r = mid-1;else
                l = mid+;
        }
        cout<<l<<endl;
    }
    return ;
}