poj3685 Matrix

思路：

二分套二分。

矩阵在每一列上是严格递增的，可以利用这一点进行二分。

实现：

 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 using namespace std;
 typedef long long ll;
 const ll INF = 0x3f3f3f3f3f3f3f3f;
 ll n, m;
 ll cal(ll i, ll j)
 {
     return i * i + i *  + j * j - j *  + i * j;
 }
 bool check(ll x)
 {
     ll sum = ;
     for (int i = ; i <= n; i++)
     {
         int l = , r = n, ans = ;
         while (l <= r)
         {
             int m = l + r >> ;
             ll tmp = cal(m, i);
             if (tmp <= x) { ans = m; l = m + ; }
             else r = m - ;
         }
         sum += ans;
     }
     return sum >= m;
 }
 int main()
 {
     int T;
     scanf("%d", &T);
     while (T--)
     {
         scanf("%lld %lld", &n, &m);
         ll l = INF, r = -INF, ans = -INF;
         for (int i = ; i <= n; i++)
         {
             l = min(l, cal(, i));
             r = max(r, cal(n, i));
         }
         while (l <= r)
         {
             ll m = l + r >> ;
             if (check(m)) { ans = m; r = m - ; }
             else l = m + ;
         }
         printf("%lld\n", ans);
     }
     return ;
 }