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) *  + i * j;
 }

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

         return sum >= m;

 }

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

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

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

       printf("%I64d\n",l);
 }

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

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

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

         return ;
 }