标签：als bool nbsp miss row span mon memory hose
| Time Limit: 6000MS | Memory Limit: 65536K | |
| Total Submissions: 7034 | Accepted: 2071 |
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
//和上一题差不多，当j固定时表达式随i的增大而增大，先二分第k小数的值s，然后枚举j， //找小于等于s的i的个数有几个，用这个值与k比较来二分。 #include<iostream> #include<cstdio> #include<cstring> using namespace std; typedef long long ll; int t; ll n,k; ll make(ll i,ll j){ return i*i+j*j+i*100000-j*100000+i*j; } bool solve(ll m){ ll cnt=0; for(int j=1;j<=n;j++){ int l=1,r=n,ans=0; while(l<=r){ int i=(l+r)>>1; if(make(i,j)<=m){ ans=i; l=i+1; }else r=i-1; } cnt+=ans; } return cnt>=k; } int main() { scanf("%d",&t); while(t--){ scanf("%lld%lld",&n,&k); ll l=-100000*n; ll r=n*n+n*n+100000*n+n*n,ans; while(l<=r){ ll m=(l+r)>>1; if(solve(m)){ ans=m; r=m-1; }else l=m+1; } printf("%lld\n",ans); } return 0; }
标签：als bool nbsp miss row span mon memory hose
原文地址：http://www.cnblogs.com/--ZHIYUAN/p/7228335.html
