[CF920G]List Of Integers

Description:

\(t\)组询问，求第\(k\)个大于\(x\)且与\(p\)互质的数

Hint:

\(x,k,p<=1e6,t<=30000\)

Solution:

推出式子后，由于要求第k大，二分答案就好了

#include<bits/stdc++.h>
using namespace std;
const int mxn=1e6+5;
int tot,vis[mxn],mu[mxn],sum[mxn],p[mxn];void sieve(int lim)
{mu[1]=1;for(int i=2;i<=lim;++i) {if(!vis[i]) mu[i]=-1,p[++tot]=i;for(int j=1;j<=tot&&p[j]*i<=lim;++j) {vis[p[j]*i]=1;if(i%p[j]==0) {mu[p[j]*i]=0;break;}mu[p[j]*i]=-mu[i];}}for(int i=1;i<=lim;++i) sum[i]=sum[i-1]+mu[i];
}int check(int x,int p)
{int ans=0;for(int i=1;i<=sqrt(p);++i) if(p%i==0) {ans+=mu[i]*(x/i);if(i*i!=p) ans+=mu[p/i]*(x/(p/i)); //这里很重要，直接将O(n)的check优化到了O(√n)}return ans;
}int main()
{int t,x,p,k;scanf("%d",&t); sieve(1000000);while(t--) {scanf("%d%d%d",&x,&p,&k);int l=x+k-1,r=9e6+5;while(l<r) {int mid=(l+r)>>1;if(check(mid,p)-check(x,p)>=k) r=mid;else l=mid+1;}printf("%d\n",r);}return 0;
}

转载于:https://www.cnblogs.com/list1/p/10372063.html

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[CF920G]List Of Integers

Description:

Hint:

Solution:

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