poj3685 Matrix

Posted 2020-10-22 王宜鸣

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：

 1 #include <cstdio>
 2 #include <cmath>
 3 #include <algorithm>
 4 using namespace std;
 5 typedef long long ll;
 6 const ll INF = 0x3f3f3f3f3f3f3f3f;
 7 ll n, m;
 8 ll cal(ll i, ll j)
 9 {
10     return i * i + i * 100000 + j * j - j * 100000 + i * j;
11 }
12 bool check(ll x)
13 {
14     ll sum = 0;
15     for (int i = 1; i <= n; i++)
16     {
17         int l = 1, r = n, ans = 0;
18         while (l <= r)
19         {
20             int m = l + r >> 1;
21             ll tmp = cal(m, i);
22             if (tmp <= x) { ans = m; l = m + 1; }
23             else r = m - 1;
24         }
25         sum += ans;
26     }
27     return sum >= m;
28 }
29 int main()
30 {
31     int T;
32     scanf("%d", &T);
33     while (T--)
34     {
35         scanf("%lld %lld", &n, &m);
36         ll l = INF, r = -INF, ans = -INF;
37         for (int i = 1; i <= n; i++)
38         {
39             l = min(l, cal(1, i));
40             r = max(r, cal(n, i));
41         }
42         while (l <= r)
43         {
44             ll m = l + r >> 1;
45             if (check(m)) { ans = m; r = m - 1; }
46             else l = m + 1;
47         }
48         printf("%lld\n", ans);
49     }
50     return 0;
51 }