BZOJ2986 Non-Squarefree Numbers

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我们可以考虑二分答案，那么考虑如何求n以内的non-squarefree numbers有多少个

考虑容斥，个数为n减一定是1个质数的平方的倍数的个数加一定是两个质数的平方的倍数的个数减……

那么就是sigma n/(i*i)*mu[i]

mu为莫比乌斯函数

这样的话i枚举到根号n就行了

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<ctime>#include<algorithm>#include<iomanip>#include<cmath>#include<vector>#include<queue>#include<stack>#include<bitset>#include<set>#include<map>using namespace std;#define MAXN 1000010#define MAXM 1010#define INF 1000000000#define eps 1e-8#define MOD 1000000007#define ll long longbool np[MAXN];int p[MAXN],tot;int u[MAXN];ll ans,n;void su(){int i,j;u[1]=1;for(i=2;i<MAXN;i++){if(!np[i]){p[++tot]=i;u[i]=-1;}for(j=1;p[j]*i<MAXN&&j<=tot;j++){np[i*p[j]]=1;if(!(i%p[j])){break;}u[i*p[j]]=-u[i];}}}ll cal(ll x){ll i;ll re=x;for(i=1;i*i<=x;i++){re-=x/(i*i)*u[i];}return re;}int main(){su();scanf("%lld",&n);ll l=0,r=210000000000ll;while(l<=r){ll mid=l+r>>1;if(cal(mid)>=n){ans=mid;r=mid-1;}else{l=mid+1;}}printf("%lld\n",ans);return 0;}/**/

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