poj3685（嵌套二分）

Matrix

Time Limit: 6000MS		Memory Limit: 65536K
Total Submissions: 4658		Accepted: 1189

Description

Given a N × N matrix A, whose element in the i-th row and j-th column A_ij 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

Source

POJ Founder Monthly Contest – 2008.08.31, windy7926778 

首先打個表看看，初步認爲左下到右上遞增。

認真一看，又不是特別有規律，在N比較大時候，遞增就木有了。

但是每一列的單調性是可以保持的。這個分別將i,j看成常數求一下導數就非常容易知道了。

這個二分有意思。

在long long 範圍內二分一個數X，>號即爲 滿足X大於矩陣的數大於等於M個

而大於矩陣的數的個數可以通過每一列二分來確定。

時間複雜度log(10^18)*N*log(N)。

#include <iostream>

using namespace std;
long long N, M;
const long long INF = 1LL << 50;
long long mtr ( long long  i, long long j )
{
	return i * i + 100000 * i + j * j - 100000 * j + i * j;
}

bool b_s ( long long X )
{
	long long res = 0;

	for ( int i = 1; i <= N; i++ )
		{
			int cnt = N ;
			int l = 1, r = N;

			while ( l <= r )
				{
					int mid = ( r + l ) >> 1;

					if ( mtr ( mid, i ) >= X )
						{
							r = mid - 1;
							cnt = mid - 1;
						}
					else
						{
							l = mid + 1;
						}
				}

			res += cnt ;
		}
	return res >= M;
}
int main()
{
	int n;
	cin >> n ;
	while ( n-- )
		{
			cin >> N >> M;
			long long l = -INF, r = INF;
			long long ans=-1;

			while ( l <= r )
				{
					long long mid = ( r + l )>>1;
					if ( b_s ( mid ) )
						{
							r = mid - 1;
							ans = mid - 1;
						}
					else
						{
							l = mid + 1;
						}
				}

			cout <<ans << endl;
		}

	return 0;
}