poj3685 Matrix (二分套二分)

Matrix

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

思路:

二分第m大的数，但是check不好搞
可以发现当 j 固定的时候 i 越大数值越大，对于每一列 j 二分求出小于等于mid的数，累加起来
如果累加的和大于等于m则说明mid大于等于第k大

code:

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
typedef long long ll;
ll n,m;
ll cal(ll i,ll j){//i大值大，j大值小
    return i*i+100000*i+j*j-100000*j+i*j;
}
bool check(ll x){//判断x是否大于等于第m大
    ll sum=0;//小于等于x的数的个数
    for(int j=1;j<=n;j++){//求出每一列比x小的数的个数累加，结果就是矩阵中小于x的数的个数
        int l=1,r=n;//二分枚举i
        int ans=0;//每一列j固定，数值根据i递增
        while(l<=r){
            int mid=(l+r)/2;
            if(cal(mid,j)<=x){
                ans=mid;
                l=mid+1;
            }else{
                r=mid-1;
            }
        }
        sum+=ans;//累加上当前列小于等于x的数的个数
    }
    return sum>=m;//x是否大于等于第m大
}
signed main(){
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%lld%lld",&n,&m);
        ll l=cal(0,n),r=cal(n,0);
        ll ans=0;
        while(l<=r){
            ll mid=(l+r)/2;
            if(check(mid)){
                ans=mid;
                r=mid-1;
            }else{
                l=mid+1;
            }
        }
        printf("%lld\n",ans);
    }
    return 0;
}