VIJOS 1889 天真的因数分解(莫比乌斯反演，容斥原理)

https://vijos.org/p/1889

同BZOJ2440..，不过这题要求的是有因数因子的，所以莫比乌斯函数要稍微改一下

 #include<algorithm>
 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<iostream>
 #define N 200000
 #define ll long long
 int mul[N+],mark[N+],p[N+];
 ll K;
 void init(){
     mul[]=;
     for (int i=;i<=N;i++){
         if (!mark[i]){
             p[++p[]]=i;
             mul[i]=;
         }
         for (int j=;j<=p[]&&p[j]*i<=N;j++){
             mark[i*p[j]]=;
             if (i%p[j]) mul[i*p[j]]=-mul[i];
             else{
                 mul[i*p[j]]=;
                 break;
             }
         }
     }
 }
 ll cal(ll x){
     ll sum=,t=sqrt(x);
     for (ll i=;i<=t;i++){
         sum+=x/(i*i)*mul[i];
     }
     return sum;
 }
 int main(){
     scanf("%lld",&K);
     init();
     ll l=K,r=25505460948LL,ans=;
     while (l<=r){
         ll mid=(l+r)>>;
         if (cal(mid)>=K) ans=mid,r=mid-;
         else l=mid+;
     }
     printf("%lld\n",ans);
 }