Codeforces-984D - XOR-pyramid - 思维

题解链接：

https://www.lucien.ink/archives/212/

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题目链接：

http://codeforces.com/contest/984/problem/D

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题目：

For an array b b of length m m we define the function f f as

f(b){b[1],f(b[1]⊕b[2],b[2]⊕b[3],…,b[m−1]⊕b[m])if m=1otherwise(1) (1) f ( b ) { b [ 1 ] , i f   m = 1 f ( b [ 1 ] ⊕ b [ 2 ] , b [ 2 ] ⊕ b [ 3 ] , … , b [ m − 1 ] ⊕ b [ m ] ) o t h e r w i s e

where ⊕ ⊕ is bitwise exclusive OR.

For example, f(1,2,4,8)=f(1⊕2,2⊕4,4⊕8)=f(3,6,12)=f(3⊕6,6⊕12)=f(5,10)=f(5⊕10)=f(15)=15 f ( 1 , 2 , 4 , 8 ) = f ( 1 ⊕ 2 , 2 ⊕ 4 , 4 ⊕ 8 ) = f ( 3 , 6 , 12 ) = f ( 3 ⊕ 6 , 6 ⊕ 12 ) = f ( 5 , 10 ) = f ( 5 ⊕ 10 ) = f ( 15 ) = 15

You are given an array a a and a few queries. Each query is represented as two integers l l and r r . The answer is the maximum value of f f on all continuous subsegments of the array al,al+1,…,ar a l , a l + 1 , … , a r .

Input

The first line contains a single integer n (1≤n≤5000) n   ( 1 ≤ n ≤ 5000 ) — the length of a a .

The second line contains nn integers a1,a2,…,an (0≤ai≤230−1) a 1 , a 2 , … , a n   ( 0 ≤ a i ≤ 2 30 − 1 ) — the elements of the array.

The third line contains a single integer q (1≤q≤100 000) q   ( 1 ≤ q ≤ 100   000 ) — the number of queries.

Each of the next q q lines contains a query represented as two integers l,r (1≤l≤r≤n) l , r   ( 1 ≤ l ≤ r ≤ n ) .

Output

Print q q lines — the answers for the queries.

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题意：

  定义一个函数 f(b) f ( b ) ， b b 对应一个数组，作用方式见题意。给你一个长度为n的序列b，然后给你q次询问，对于每次询问，给你l r，问你在 [l,r] [ l , r ] 这段区间内所有子串中 f f 的最大值为多少。

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思路：

  我开始看这道题的时候还剩20分钟了，然而小伙伴已经思考了很久了…并且发现了一个十分关键的性质。

  定义 f[l][r] f [ l ] [ r ] 为在 [l,r] [ l , r ] 这个区间内函数 f f 的结果，那么有：

f[l][r]{b[l],f[l][r−1]⊕f[l+1][r]if l=rotherwise(2) (2) f [ l ] [ r ] { b [ l ] , i f   l = r f [ l ] [ r − 1 ] ⊕ f [ l + 1 ] [ r ] o t h e r w i s e

  于是我们可以根据这个性质 O(n2) O ( n 2 ) 预处理出所有区间的 f f ，与此同时可以把答案离线下来，然后 O(1) O ( 1 ) 出答案即可。

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实现：

#include 
int f[5007][5007], ans[5007][5007], n, q, l, r, i;
int main() {
    for (scanf("%d", &n), i = 1; i <= n; i++) scanf("%d", f[i] + i), ans[i][i] = f[i][i];
    for (int len = 2; len <= n; len++)
        for (l = 1, r; (r = l + len - 1) <= n; l++) {
            f[l][r] = f[l][r - 1] ^ f[l + 1][r];
            ans[l][r] = std::max(f[l][r], std::max(ans[l][r - 1], ans[l + 1][r]));
        }
    for (scanf("%d", &q), i = 0; i < q; i++) {
        scanf("%d%d", &l, &r);
        printf("%d\n", ans[l][r]);
    }
    return 0;
}