省选专练POI2015Kwadraty

最新推荐文章于 2024-04-13 18:18:43 发布

LauZiyang

最新推荐文章于 2024-04-13 18:18:43 发布

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分类专栏： 2015 智商题 POI

本文链接：https://blog.youkuaiyun.com/fcb_x/article/details/80210337

POI 同时被 3 个专栏收录

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2015

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智商题

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本文探讨了一种名为“毒瘤打表”的特殊算法，并详细解释了其背后的数学原理及其实现过程。通过分析506这个关键数字，文章展示了如何高效地解决特定类型的问题，并给出了一份详细的代码实现。

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毒瘤

打表找规律

为什么Claris大神选择了506

原因是对于506以上的

绝对单调

第二问不必统计答案

#include<bits/stdc++.h>
const int N=507;
using namespace std;
typedef long long ll;
ll n,l=12,r=1442250,mid,t,ans;
int i,j,v[N]={},sum[N]={};
int f[N]={0,1,0,0,2,2,0,0,0,3,3,0,0,3,3,0,4,4,0,0,4,4,0,0,0,4,4,0,0,4,4,0,0,0,5,5,6,6,5,5,6,5,5,0,0,5,5,0,0,6,5,5,6,6,5,5,6,6,7,7,0,6,6,7,8,6,6,0,8,7,6,6,0,8,6,6,0,6,6,7,8,6,6,7,7,7,6,6,7,7,6,6,0,8,7,7,0,9,7,7,7,7,7,7,7,7,7,9,0,8,7,7,0,8,7,7,8,8,8,7,7,8,8,7,7,8,7,7,0,8,7,7,9,8,8,7,7,9,8,7,7,8,8,8,9,8,8,8,8,8,8,8,8,8,8,8,9,10,8,8,9,9,8,8,8,8,8,8,8,8,8,9,9,10,8,8,9,10,8,8,9,9,9,8,8,9,9,8,8,10,8,8,9,10,8,8,9,9,9,8,8,9,9,8,8,9,9,9,9,10,9,9,9,10,9,9,9,9,10,9,9,9,9,9,9,10,9,9,9,9,9,9,9,9,9,9,9,10,10,9,9,10,10,9,9,9,9,9,9,9,9,9,10,10,10,9,9,11,10,9,9,10,10,10,9,9,10,10,9,9,10,9,9,11,10,9,9,11,10,10,9,9,10,10,9,9,10,10,10,11,10,10,10,11,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,11,10,10,10,11,10,10,10,10,11,10,10,10,10,10,10,11,10,10,10,10,10,10,10,10,10,10,10,11,11,10,10,11,11,10,10,10,10,10,10,10,10,10,11,11,11,10,10,11,11,10,10,11,11,11,10,10,11,11,10,10,11,10,10,11,11,10,10,11,12,11,10,10,11,11,10,10,11,11,11,11,11,11,11,11,12,11,11,11,12,11,11,11,11,11,11,11,11,11,11,11,12,11,11,11,12,11,11,11,11,11,11,11,11,11,11,11,11,11,11,11,12,11,11,11,12,11,11,11,11,12,11,11,11,11,11,11,12,11,11,11,11,11,11,11,11,11,11,11,12,12,11,11,12,12,11,11,11,11,11,11,11,11,11,12,12,12,11,11,12,12,11,11,12,12,12,11,11,12,12,11,11,12,11,11,12,12,11,11,12,12,12,11,11,12,12,11,11};
ll F(ll x){
	return x*(x+1)*(x*2+1)/6;
}
int main(){
	scanf("%lld",&n);
  	for(i=2;i<N;i++)
  		if(f[i])
		  	for(j=1;j<i;j++)
			  	if(!f[j]||f[j]>f[i])
					v[j]=1;
  	for(i=2;i<N;i++)
	  	sum[i]=sum[i-1]+v[i];
 	if(n<N){
    	if(f[n])
			printf("%d",f[n]);
		else 
			putchar('-');
    	printf(" %d",sum[n]);
    	return 0;
  	}
	while(l<=r){
  		mid=(l+r)/2;
  		if(F(mid)>=n){
  			t=mid;
  			r=mid-1;
  		}
  		else l=mid+1;
  	}
//  	while(l<=r)if(F(mid=(l+r)>>1)>=n)r=(t=mid)-1;else l=mid+1;
//  	printf("%lld ",t+);
  	if((F(t)>n&&F(t)-n<=128&&!f[F(t)-n])){
  		cout<<t+1;
  	}
  	else cout<<t;
//  	printf("%lld ",t+(F(t)>n&&F(t)-n<=128&&!f[F(t)-n]));
  	cout<<" ";
	for(ans=(t-12)*31+sum[N-1],i=1;i<=128;i++)
		if(!f[i]&&F(t)-i<=n)
			ans++;
  	printf("%lld",ans);
  	return 0;
}