poj 3685 Matrix（二分搜索之查找第k大的值）

Description

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

 

Source

POJ Founder Monthly Contest – 2008.08.31, windy7926778 

 

 题目大意：题目意思很简单。这个题目乍一看，先打n为比较小例如8的表，会觉得很有规律，大小规律是从右上往左下依次增大，但是这个规律到n为5*10^4就不一定保持了。

           解题思路：有一个规律是看得见的，j不变i增大函数值也在增大。根据这个可以对这n列二分得到<x的值，同样所求的x也是可以二分慢慢靠近最后的结果，我处理得到最后的结果是个数为m+1的最小值，所以最后mid-1即为答案。

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<math.h>
#include<stdlib.h>
using namespace std;
#define inf 1<<30
#define ll long long
#define N 50006
ll n,m;
ll cal(ll i,ll j){
    return i*i+100000*i+j*j-100000*j+i*j;
}
bool solve(ll mid){
    ll cnt=0;
    for(ll j=1;j<=n;j++){
        ll low=1;
        ll high=n+1;
        while(low<high){
            ll tmp=(low+high)>>1;//另外的写法
            ll ans=cal(tmp,j);
            if(ans>=mid){
                high=tmp;
            }
            else{
                low=tmp+1;
            }
        }
        cnt+=low-1;//另外的写法
    }
    if(cnt>=m) return true;
    return false;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--){

        scanf("%I64d%I64d",&n,&m);
        ll low=-1e12;
        ll high=1e12;

        while(low<high){
            ll mid=(low+high)>>1;//另外的写法
            if(solve(mid)){
                high=mid;
            }
            else{
                low=mid+1;
            }
        }
        printf("%I64d\n",low-1);//另外的写法

    }
    return 0;
}

 

 

另外的写法，试比较

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<math.h>
#include<stdlib.h>
using namespace std;
#define inf 1<<30
#define ll long long
#define N 50006
ll n,m;
ll cal(ll i,ll j){
    return i*i+100000*i+j*j-100000*j+i*j;
}
bool solve(ll mid){
    ll cnt=0;
    for(ll j=1;j<=n;j++){
        ll low=1;
        ll high=n+1;
        ll tmp=(low+high)>>1;
        while(low<high){
            ll ans=cal(tmp,j);
            if(ans>=mid){
                high=tmp;
            }
            else{
                low=tmp+1;
            }
            tmp=(low+high)>>1;
        }
        cnt+=tmp-1;
    }
    if(cnt>=m) return true;
    return false;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--){

        scanf("%I64d%I64d",&n,&m);
        ll low=-1e12;
        ll high=1e12;
        ll mid=(low+high)>>1;
        while(low<high){
            if(solve(mid)){
                high=mid;
            }
            else{
                low=mid+1;
            }
            mid=(low+high)>>1;
        }
        printf("%I64d\n",mid-1);

    }
    return 0;
}