POJ3685 Matrix 二分搜索

        題目大意：給定NXN的矩陣,第i行第j列的值Aij =  i² + 100000 × i + j² - 100000 × j + i × j, N不超過5萬。求第M小的數。

        這題仍然是用二分搜索,設C(X):=矩陣中值不小於X的個數不少於K。至於如何判斷命題爲真，可以發現這個等式的值是逐行單調遞增的，所以可以對每一列再進行二分搜索，找出不小於X的個數就可以了。

#include <stdio.h>
#include <vector>
#include <math.h>
#include <string.h>
#include <string>
#include <iostream>
#include <queue>
#include <list>
#include <algorithm>
#include <stack>
#include <map>

using namespace std;

bool CC(long long x, int n, long long k)
{
	long long curcount = 0;

	int l = -1;
	for (int i = 1; i <= n; i++)
	{
		
		l = 0;
		int r = n + 1;
		while (r - l > 1)
		{
			long long mid = (r + l) / 2;
			if (mid * mid + 100000 * mid + i * i - 100000 * i + mid * i >= x)
			{
				r = mid;
			}
			else
				l = mid;
		}

		curcount += (n - r + 1);


		if (curcount >= k)
		{
			return true;
		}
	}
	if (curcount < k)
	{
		return false;
	}
	else
		return true;
}

int main()
{
#ifdef _DEBUG
	freopen("d:\\in.txt", "r", stdin);
#endif
	int k;
	scanf("%d", &k);
	for (int i = 0; i < k; i++)
	{
		long long n, m;
		scanf("\n%I64d %I64d\n", &n, &m);
		//qsort(values, n, sizeof(int), compp);
		long long l = 50000 * (long long)50000 - 100000LL * 50000;
		long long r = 50000LL * 50000LL * 3 + 1;
		
		long long k = (long long)n * n - m + 1;
		while (r - l > 1)
		{
			long long mid = (r + l) / 2;
			if (CC(mid, n, k))
			{
				l = mid;
			}
			else
				r = mid;
		}
		printf("%I64d\n", l);
	}
	return 1;
}