CF983B XOR-pyramid

思路：

简单的区间dp。
实现：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int MAXN = 5005;
 4 int a[MAXN][MAXN];
 5 int main()
 6 {
 7     int n, q, l, r;
 8     while (cin >> n)
 9     {
10         memset(a, 0, sizeof a);
11         for (int i = 1; i <= n; i++) cin >> a[i][i];
12         for (int i = 2; i <= n; i++)
13         {
14             for (int j = 1; j <= n - i + 1; j++)
15             {
16                 a[j][j + i - 1] = a[j][j + i - 2] ^ a[j + 1][j + i - 1];
17             }
18         }
19         for (int i = 2; i <= n; i++)
20         {
21             for (int j = 1; j <= n - i + 1; j++)
22             {
23                 a[j][j + i - 1] = max(a[j][j + i - 1], max(a[j][j + i - 2], a[j + 1][j + i - 1]));
24             }
25         }
26         cin >> q;
27         while (q--)
28         {
29             cin >> l >> r;
30             cout << a[l][r] << endl;
31         }
32     }
33     return 0;
34 }

 

转载于:https://www.cnblogs.com/wangyiming/p/9060310.html