标签：scanf ima segment lin def operation otherwise amp main
设f[l,r]为区间[l,r]的f值，那么可以很容易的发现f[l,r] = f[l+1,r] ^ f[l,r-1] (l<r) or a[l] (l==r)，这个在纸上画一画就发现了。
于是就成了一个SB题了2333
Discription
For an array bb of length mm we define the function ff as
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)=15f(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 aa and a few queries. Each query is represented as two integers ll and rr. The answer is the maximum value of ff on all continuous subsegments of the array al,al+1,…,aral,al+1,…,ar.
Input
The first line contains a single integer nn (1≤n≤50001≤n≤5000) — the length of aa.
The second line contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤230?10≤ai≤230?1) — the elements of the array.
The third line contains a single integer qq (1≤q≤1000001≤q≤100000) — the number of queries.
Each of the next qq lines contains a query represented as two integers ll, rr (1≤l≤r≤n1≤l≤r≤n).
Output
Print qq lines — the answers for the queries.
Examples
3
8 4 1
2
2 3
1 2
5
12
6
1 2 4 8 16 32
4
1 6
2 5
3 4
1 2
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].
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=5005;
int f[maxn][maxn],ans[maxn][maxn],n,Q,L,R;
int main(){
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%d",&f[i][i]);
for(int l=1;l<n;l++)
for(int i=1,j=l+1;j<=n;i++,j++){
f[i][j]=f[i+1][j]^f[i][j-1];
ans[i][j]=max(f[i][j],max(ans[i+1][j],ans[i][j-1]));
}
scanf("%d",&Q);
while(Q--) scanf("%d%d",&L,&R),printf("%d\n",ans[L][R]);
return 0;
}
标签：scanf ima segment lin def operation otherwise amp main
原文地址：https://www.cnblogs.com/JYYHH/p/9058066.html
