poj 2773 Happy 2006 容斥原理+二分-优快云博客

本文介绍了如何使用容斥原理找出与给定整数n互质的数中第k个数。通过算法实现，解决了一个在数论中的常见问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
vector<int> v;
void init(int n) {
    v.clear();
    for(int i = 2; i*i<=n; i++) {
        if(n%i==0) {
            v.pb(i);
        }
        while(n%i==0)
            n/=i;
    }
    if(n!=1)
        v.pb(n);
}
ll solve(ll x) {
    int len = v.size();
    ll ret = 0;
    for(int i = 1; i<(1<<len); i++) {
        ll cnt = 0, ans = 1;
        for(int j = 0; j<len; j++) {
            if((1<<j)&i) {
                cnt++;
                ans *= v[j];
            }
        }
        if(cnt&1) {
            ret += x/ans;
        } else {
            ret -= x/ans;
        }
    }
    return x-ret;
}
int main()
{
    int n, m;
    while(cin>>n>>m) {
        init(n);
        ll l = 0, r = 1LL<<61, ans;
        while(r>=l) {
            ll mid = (l+r)/2;
            ll ret = solve(mid);
            if(ret>=m) 
                r = mid-1;else
                l = mid+1;
        }
        cout<<l<<endl;
    }
    return 0;
}

转载于:https://www.cnblogs.com/yohaha/p/5082996.html