Matrix POJ - 3685 + binary enumeration

Portal: https://vjudge.net/contest/363330#problem/A

 

Meaning of the questions, nxn matrix, each column is increasing, seeking what the entire matrix of small to large m-th number is

Error-prone Point: binary algorithm is to find a number x, which is smaller than a number of m, so that a reduction on the x and the like in the m-th number.

If you find a direct result res-half, smaller than the number of res have m-1 Ge, but then if m = 1, it will cause a dichotomous outcomes not the least that number.

 

Then half of each column when the idea is that, first find a position greater than or equal to k, is the current position of -1 to accumulate results

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define maxn 50005
#define mod 1000000000
#define INF 0x3f3f3f3f
using namespace std; 
typedef long long ll;
ll n,m,ans;
 
ll f(ll x,ll y)
{
    return x * x + 100000 * x + y * y- 100000 * y + x * y;
}
ll calnum(ll k)
{
    ll i,j;
    Le LL, RI, MID, SUM = 0 ;
     for (J = . 1 ; J <= n-; J ++)   // for each column enum 
    {
        the = 1 ;
        RI = n-+ . 1 ;
         the while (Le <RI)    // Get the number k is less than 
        {
            mid=(le+ri)>>1;
            if(f(mid,j)>=k) ri=mid;
            else le=mid+1;
        }
        sum + = to- 1 ;
    }
    return sum;
}
void solve()
{
    ll le,mid,ri,t;
    ri=1LL<<50;
    Le = - RI;
     the while (Le <RI)   // half answers 
    {
        mid=(le+ri)>>1;
        t=calnum(mid);
        if(t>=m) ri=mid;
        else le=mid+1;
    }
    years = le 1 ;
}
int main ()
{
    int i,j,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld%lld",&n,&m);
        solve();
        printf("%lld\n",ans);
    }
    return 0;
}