bzoj 2986: Non-Squarefree Numbers （容斥原理）

题目描述

传送门

题解

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define LL long long
#define N 1000000
using namespace std;
LL n;
int pd[N+3],prime[N+3],mu[N+3];
void init()
{
mu[1]=1;
for (int i=2;i<=N;i++) {
if (!pd[i]) {
prime[++prime[0]]=i;
mu[i]=-1;
}
for (int j=1;j<=prime[0];j++) {
if (i*prime[j]>N) break;
pd[i*prime[j]]=1;
if (i%prime[j]==0) {
mu[i*prime[j]]=0;
break;
}
mu[i*prime[j]]=-mu[i];
}
}
}
LL check(LL x)
{
LL ans=0;
for (LL i=2;i*i<=x;i++)
if (mu[i]) ans=ans-mu[i]*(x/(i*i));
return ans;
}
int main()
{
//  freopen("a.in","r",stdin);
//  freopen("my.out","w",stdout);
init();
scanf("%lld",&n);
LL l=0; LL r=30000000000LL; LL ans=r;
while (l<=r) {
LL mid=(l+r)/2;
if (check(mid)>=n) ans=min(ans,mid),r=mid-1;
else l=mid+1;
}
printf("%lld\n",ans);
}