[模拟+二分]zoj 3470：Magic Squares

大致题意：

    如题目中给出的图片

对于这样的一个无线扩展出去的图，输入一个数n，求出数字上下左右的4个数字，按造升序输出。

 

大致思路：

    突破点在，对于每一圈右下角的数字都是（a*2-1）*（a*2-1），a为当前在第a圈。如此，通过二分枚举判定出这个点在第几个圈内。然后在推导这个点和上下左右点的关系。

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
long long getround(long long n){
	long long left=0,mid,right=sqrt(2000000003),res;//1000000003
	if(n==1)return 1;
	while(right>=left){
		mid=(right+left)/2;
		long long tmp=mid;
		tmp=tmp*2-1;
		tmp*=tmp;
		if(tmp<n){
			res=mid;
			left=mid+1;
		}
		else{
			right=mid-1;
		}
	}
	return res+1;
}
int main(){
	long long i,j,k,loc,a,s,l,cas,b,c,d,n;
	cin>>cas;
	while(cas--){
		cin>>n;
		loc=getround(n);
		if(n==1){
			printf("2 4 6 8\n");
			continue;
		}
		l=loc*2-1;
		s=(loc-1)*2-1;
		s*=s;
		if(n==s+1){
			cout<<s<<" "<<n+1<<" "<<(loc*2-1)*(loc*2-1)<<" "<<(loc*2-1)*(loc*2-1)+2<<endl;

			continue;
		}
		if(n>s+1&&n<=s+l-2){
			a=s-3*(l-3)+1;
			a-=(s+l-1-n);
			b=n-1;
			c=n+1;
			d=(l-1)*4+1+n;
			cout<<a<<" "<<b<<" "<<c<<" "<<d<<endl;
			continue;
		}
		if(n==s+l-1){
			a=n-1;
			b=n+1;
			c=n+4*l-3;
			d=c+2;
			cout<<a<<" "<<b<<" "<<c<<" "<<d<<endl;
		}
		if(n>s+l-1&&n<s+2*(l-1)){
			a=n+1-4*(l-2);
			b=n-1;
			c=n+1;
			d=n+4*l-1;
			cout<<a<<" "<<b<<" "<<c<<" "<<d<<endl;
			continue;
		}
		if(n==s+2*(l-1)){
			a=n-1;
			b=n+1;
			c=n+4*l-1;
			d=n+4*l+1;
			cout<<a<<" "<<b<<" "<<c<<" "<<d<<endl;
			continue;
		}
		if(n>s+2*(l-1)&&n<s+3*(l-1)){
			a=n-1-4*(l-2);
			b=n-1;
			c=n+1;
			d=n+4*l+1;
			cout<<a<<" "<<b<<" "<<c<<" "<<d<<endl;
			continue;
		}
		if(n==s+3*(l-1)){
			a=n-1;
			b=n+1;
			c=n+4*l+1;
			d=n+4*l+3;
			cout<<a<<" "<<b<<" "<<c<<" "<<d<<endl;
			continue;
		}
		if(n>s+3*(l-1)&&n<s+4*(l-1)){
			a=n-3-4*(l-2);
			b=n-1;
			c=n+1;
			d=n+4*l+3;
			cout<<a<<" "<<b<<" "<<c<<" "<<d<<endl;
			continue;
		}
		if(n==s+4*(l-1)){
//		    cout<<"fuck";
			a=n-3-4*(l-2);
			b=n-1;
			c=n+1;
			d=n+4*l+3;
			cout<<a<<" "<<b<<" "<<c<<" "<<d<<endl;
		}
	}
	return 0;
}