POJ 3685 Matrix 二分答案里面套二分

Given a N × N matrix A, whose element in the i-th row and j-th column Aij is an number that equals i2 + 100000 × i + j2 - 100000 × j + i × j, you are to find the M-th smallest element in the matrix.

Input

The first line of input is the number of test case.
For each test case there is only one line contains two integers, N(1 ≤ N ≤ 50,000) and M(1 ≤ M ≤ N × N). There is a blank line before each test case.

Output

For each test case output the answer on a single line.

Sample Input

12

1 1

2 1

2 2

2 3

2 4

3 1

3 2

3 8

3 9

5 1

5 25

5 10

Sample Output

3
-99993
3
12
100007
-199987
-99993
100019
200013
-399969
400031
-99939

第一眼看到这个题目，打了个表，wa了一万年，打表那个规律是i越大，j越小  ，值越大，其实看那个表达式，i越大值越大，但是

当i不同时，j越大值不一定越大 所以那个规律是错的

正解是 

当j相同时 i越大值越大，  当i相同时，j越大值越小

所以

二分答案，在二分答案时再枚举j，二分求出每列<=二分答案的值的数量，将所有列的<=二分答案的值的数量加起来，假设用num计数，num就是二分答案时第num大数，

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstdlib>
#include<deque>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int>P;
const int len=1e5+5;
const ll mod=998244353;
const ull seed=131;
ll solve(ll i,ll j)
{
	return i*i+100000*i+j*j-100000*j+i*j;
}
ll n,k;
int judge(ll m)
{
	ll num=0;
	for(ll i=1;i<=n;++i)
	{
		ll l=0,r=n,mid;//边界问题也要注意 
		while(l<r)
		{
			mid=(l+r+1)/2;
			if(solve(mid,i)<=m)l=mid;
			else r=mid-1;
		} 
		num+=l;
	}
	return num>=k; 
}
int main()
{	
	int t;
	cin>>t;
	while(t--)
	{
		scanf("%lld%lld",&n,&k);
		ll l=-1e12,r=1e12,mid;
		while(l<r)
		{
			mid=(l+r)>>1;
			//注意这里，/是向0取整-5/2=-2 
			// >> 是向左取整  (-5)>>1=-3
			if(judge(mid))r=mid;
			else l=mid+1;
		}
		printf("%lld\n",r);
	}
}