BZOJ - 2440 容斥定理

组合枚举n/i/i，贡献为miu倍

/*H E A D*/
int mu[maxn],prime[maxn],cnt;
bool isprime[maxn];
void sai(int n){
    mu[0]=0;mu[1]=1;
    rep(i,2,n) isprime[i]=1;
    rep(i,2,n){
        if(isprime[i]){
            prime[++cnt]=i;
            mu[i]=-1;
        }
        rep(j,1,cnt){
            if(i*prime[j]>n) break;
            isprime[i*prime[j]]=0;
            if(i%prime[j]==0){
                mu[i*prime[j]]=0;
                break;
            }
            mu[i*prime[j]]=-mu[i];
        }
    }
}
ll C(ll x){
    ll rt=sqrt(x+0.5),ans=0;
    rep(i,1,rt){
        ans+=x/i/i*mu[i];
    }
    return ans;
}
int main(){
    sai((int)1e6-11);
    int T=read();
    while(T--){
        ll x=read();
        ll l=x,r=2e9+50,mid,ans;
        while(r>=l){
            mid=l+r>>1;
            if(C(mid)>=x){
                r=mid-1;
                ans=mid;
            }
            else{
                l=mid+1;
            }
        }
        println(ans);
    }
    return 0;
}