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;
}