[BZOJ2440][中山市选2011]完全平方数（莫比乌斯函数，二分）

题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=2440

题意：RT。

是https://wenku.baidu.com/view/fbec9c63ba1aa8114431d9ac.html中的一道例题，拿来切了练手。

mark一个博客：http://blog.csdn.net/acdreamers/article/details/8542292

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
const LL maxn = 1001000;
LL vis[maxn], mu[maxn], prime[maxn], cnt;
LL n;

void init() {
    memset(vis,0,sizeof(vis));
    mu[1] = 1;
    cnt = 0;
    for(LL i=2; i<maxn; i++) {
        if(!vis[i]) {
            prime[cnt++] = i;
            mu[i] = -1;
        }
        for(LL j=0; j<cnt&&i*prime[j]<maxn; j++) {
            vis[i*prime[j]] = 1;
            if(i%prime[j]) mu[i*prime[j]] = -mu[i];
            else {
                mu[i*prime[j]] = 0;
                break;
            }
        }
    }
}

bool check(LL x) {
    LL xx = (LL)sqrt(x);
    LL ret = 0;
    for(LL i = 1; i <= xx; i++) {
        ret += mu[i]*(LL)(x/(i*i));
    }
    return ret < n;
}

signed main() {
    // freopen("in", "r", stdin);
    init();
    LL T;
    scanf("%lld", &T);
    while(T--) {
        scanf("%lld", &n);
        LL lo = 1, hi = (LL)1e10;
        LL ret = 0;
        while(lo <= hi) {
            LL mid = (lo + hi) >> 1;
            if(check(mid)) lo = mid + 1;
            else hi = mid - 1;
        }
        printf("%lld
", lo);
    }
    return 0;
}