POJ 2773 Happy 2006 【数论，容斥原理+二分】

再奉上一篇容斥原理的题目，其实还是统计区间里与某个数互素的数的个数。

同类型题目：【HDU 1695 GCD】【HDU 4407 SUM】

这道题目只需要二分区间（1，x）的右端点x，统计（1，x）与s互素的数的个数即可。

 

#include <cstdio>

#include <cstring>

#include <vector>

#include <algorithm>

using namespace std;

typedef long long LL;

#define N 1000100

bool is[N];

vector<int> pr, g;

int kk;



void prime() {

    pr.clear();

    memset(is, true, sizeof(is));

    pr.push_back(2);

    for (int i=3; i<N; i+=2)

        if (is[i]) {

            pr.push_back(i);

            for (int j=i+i; j<N; j+=i) is[j] = false;

        }

}

bool ok(LL x, int p) {

    LL v, all = x, c, k;



    for (int s=1; s<(1<<g.size()); s++) {

        c = 0, v = 1;

        for (int i=0; i<g.size(); i++)

            if (s & (1<<i)) {

                c++;

                v *= g[i];

            }

        k = x / v;

        if (c % 2 == 1) all -= k;

        else all += k;

    }

    return all >= kk;

}

int main() {



    prime();



    int m;

    while (scanf("%d%d", &m, &kk) == 2) {

        int n = m;

        g.clear();

        for (int i=0; i<pr.size() && pr[i]<=n; i++)

            if (n % pr[i] == 0) {

                g.push_back(pr[i]);

                while (n % pr[i] == 0) n /= pr[i];

            }

        LL l = 1, r = 1e17, ans, mid;

        while (l <= r) {

            mid = (l + r) >> 1;

            if (ok(mid, m)) {

                ans = mid;

                r = mid -1;

            } else l = mid + 1;

        }

        printf("%lld\n", ans);

    }

    return 0;

}