POJ 3685

Matrix

Time Limit: 6000MS		Memory Limit: 65536K
Total Submissions: 4428		Accepted: 1102

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 

 

首先二分第m小的数，由公式易得每列的数是按照从小到大排列的，因此可二分，两次二分即可。

 

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>

using namespace std;

typedef long long ll;

int n;
ll m;

ll cal(ll i,ll j) {
        return i * i + j * j + (i - j) * 100000 + i * j;
}

bool judge(ll x) {
        ll sum = 0;
        for(int j = 1; j <= n; ++j) {
                int l = 0,r = n;
                while(l < r){
                        int mid = (l + r + 1) / 2;
                        if(cal(mid,j) <= x) {
                                l = mid;
                        } else {
                                r = mid - 1;
                        }
                }
                sum += l;
        }

        return sum >= m;

}

void solve() {
        ll l = -1e12,r = 1e12;

        //printf(" n = %d l = %lld r = %lld
",n,l,r);

        while(l < r) {
                ll mid = (l + r) >> 1;
                if(judge(mid)) r = mid;
                else l = mid + 1;
        }



      printf("%I64d
",l);
}

int main() {
       // freopen("sw.in","r",stdin);

        int t;
        scanf("%d",&t);

        while(t--) {
                scanf("%d%I64d",&n,&m);
                solve();
        }

        return 0;
}