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 +  × i + j2 -  × 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( ≤ N ≤ ,) and M( ≤ 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

Sample Output

-

-
-

-

-

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+*i+j*j-*j+i*j;
 }
 bool solve(ll mid){
     ll cnt=;
     for(ll j=;j<=n;j++){
         ll low=;
         ll high=n+;
         while(low<high){
             ll tmp=(low+high)>>;//另外的写法
             ll ans=cal(tmp,j);
             if(ans>=mid){
                 high=tmp;
             }
             else{
                 low=tmp+;
             }
         }
         cnt+=low-;//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)>>;//另外的写法
             if(solve(mid)){
                 high=mid;
             }
             else{
                 low=mid+;
             }
         }
         printf("%I64d\n",low-);//另外的写法

     }
     return ;
 }

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+*i+j*j-*j+i*j;
 }
 bool solve(ll mid){
     ll cnt=;
     for(ll j=;j<=n;j++){
         ll low=;
         ll high=n+;
         ll tmp=(low+high)>>;
         while(low<high){
             ll ans=cal(tmp,j);
             if(ans>=mid){
                 high=tmp;
             }
             else{
                 low=tmp+;
             }
             tmp=(low+high)>>;
         }
         cnt+=tmp-;
     }
     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)>>;
         while(low<high){
             if(solve(mid)){
                 high=mid;
             }
             else{
                 low=mid+;
             }
             mid=(low+high)>>;
         }
         printf("%I64d\n",mid-);

     }
     return ;
 }