D - XOR-pyramid

思路：

區間dp

dp[l][r]表示ƒ([l, r])的值

顯然，狀態轉移方程為dp[l][r] = dp[l][r-1] ^ dp[l+1][r]

初始狀態dp[i][i] = a[i]

可是，這道題求的是這段區間包含的某一連續區間的最大值

那麽用差不多的轉移方程再求一遍區間最大值：dp[l][r] = max(dp[l][r]，dp[l][r-1]，dp[l+1][r])

代碼：

#include<bits/stdc++.h>
using namespace std;
#define LL long long 
#define
 
 pb push_back
#define mem(a, b) memset(a, b, sizeof(a))

const int N = 5e3 + 5;
int dp[N][N], a[N];
int main() {
    int n, q, l, r;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = n; i >= 1; i--) {
        dp[i][i] = a[i];
        
 
for (int j = i+1; j <= n; j++) {
            dp[i][j] = dp[i][j-1] ^ dp[i+1][j];
        }
    }
    for (int i = n; i >= 1; i--) {
        for (int j = i+1; j <= n; j++) {
            dp[i][j] = max(dp[i][j], max(dp[i][j-1], dp[i+1][j]));
        }
    }
    scanf("%d", &q);
    
 
while(q--) {
        scanf("%d %d", &l, &r);
        printf("%d\n", dp[l][r]);
    }
    return 0;
}

Codeforces 984 D - XOR-pyramid