Codeforces Round #483 (Div. 2) D. XOR-pyramid

D. XOR-pyramid

time limit per test

2 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

For an array $\mathrm{bb of\; length mm we\; define\; the\; function ff as}$

$\mathrm{f(b)=\{b[1]if m=1f(b[1]\oplus b[2],b[2]\oplus b[3],\dots ,b[m-1]\oplus b[m])otherwise,f(b)=\{b[1]if m=1f(b[1]\oplus b[2],b[2]\oplus b[3],\dots ,b[m-1]\oplus b[m])otherwise,}$

where $\oplus \oplus is bitwise\; exclusive\; OR.$

For example, $\mathrm{f(1,2,4,8)=f(1\oplus 2,2\oplus 4,4\oplus 8)=f(3,6,12)=f(3\oplus 6,6\oplus 12)=f(5,10)=f(5\oplus 10)=f(15)=15f(1,2,4,8)=f(1\oplus 2,2\oplus 4,4\oplus 8)=f(3,6,12)=f(3\oplus 6,6\oplus 12)=f(5,10)=f(5\oplus 10)=f(15)=15}$

You are given an array $\mathrm{aa and\; a\; few\; queries.\; Each\; query\; is\; represented\; as\; two\; integers ll and rr.\; The\; answer\; is\; the\; maximum\; value\; of ffon\; all\; continuous\; subsegments\; of\; the\; array al,al+1,\dots ,aral,al+1,\dots ,ar.}$

Input

The first line contains a single integer $\mathrm{nn (1\le n\le 50001\le n\le 5000) —\; the\; length\; of aa.}$

The second line contains $\mathrm{nn integers a1,a2,\dots ,ana1,a2,\dots ,an (0\le ai\le 230-10\le ai\le 230-1) —\; the\; elements\; of\; the\; array.}$

The third line contains a single integer $\mathrm{qq (1\le q\le 1000001\le q\le 100000) —\; the\; number\; of\; queries.}$

Each of the next $\mathrm{qq lines\; contains\; a\; query\; represented\; as\; two\; integers ll, rr (1\le l\le r\le n1\le l\le r\le n).}$

Output

Print $\mathrm{qq lines —\; the\; answers\; for\; the\; queries.}$

Examples

input

Copy

3
8 4 1
2
2 3
1 2

output

Copy

5
12

input

Copy

6
1 2 4 8 16 32
4
1 6
2 5
3 4
1 2

output

Copy

60
30
12
3

Note

In first sample in both queries the maximum value of the function is reached on the subsegment that is equal to the whole segment.

In second sample, optimal segment for first query are $[3,6][3,6],\; for\; second\; query\; — [2,5][2,5],\; for\; third\; — [3,4][3,4],\; for\; fourth\; — [1,2][1,2].$

 

 这题就是在异或操作下的区间最大值。

 n《=5000 可以用区间DP做。

  dp[i][j] 表示长度为i 起始点是j 的区间最大值。

  

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 typedef long long LL ;
 5 const int maxn = 5e3 + 7;
 6 int dp[maxn][maxn];
 7 
 8 int main() {
 9     int n;
10     scanf("%d", &n );
11     for (int i = 1 ; i <= n ; i++)
12         scanf("%d", &dp[1][i]);
13     for (int i = 2 ; i <= n ; i++ )
14         for (int j = 1 ; j <= n - i + 1 ; j++)
15             dp[i][j] = dp[i - 1][j] ^ dp[i - 1][j + 1];
16     for (int i = 2 ; i <= n ; i++)
17         for (int j = 1 ; j <= n - i + 1 ; j++)
18             dp[i][j] = max(dp[i][j], max(dp[i - 1][j], dp[i - 1][j + 1]));
19     int q;
20     scanf("%d", &q);
21     while(q--) {
22         int x, y;
23         scanf("%d%d", &x, &y);
24         int len = y - x + 1;
25         printf("%d\n", dp[len][x]);
26     }
27     return 0;
28 }

 

转载于:https://www.cnblogs.com/qldabiaoge/p/9049908.html