题目链接:这里写链接内容
Problem Description
In Land waterless, water is a very limited resource. People always fight for the biggest source of water. Given a sequence of water sources with a1,a2,a3,…,an representing the size of the water source. Given a set of queries each containing 2 integers l and r, please find out the biggest water source between al and ar.
Input
First you are given an integer T(T≤10) indicating the number of test cases. For each test case, there is a number n(0≤n≤1000) on a line representing the number of water sources. n integers follow, respectively a1,a2,a3,…,an, and each integer is in {1,…,106}. On the next line, there is a number q(0≤q≤1000) representing the number of queries. After that, there will be q lines with two integers l and r(1≤l≤r≤n) indicating the range of which you should find out the biggest water source.
Output
For each query, output an integer representing the size of the biggest water source.
Sample Input
3
1
100
1
1 1
5
1 2 3 4 5
5
1 2
1 3
2 4
3 4
3 5
3
1 999999 1
4
1 1
1 2
2 3
3 3
Sample Output
100
2
3
4
4
5
1
999999
999999
1
区间最值查询问题，直接上st。
#include<iostream>
#include<cstdio>const int N=1008;
int a[N],stmax[N][N],log[N];
int n;
using namespace std;void init() {log[0]=-1;for(int i=1; i<=1000; i++) {log[i]=log[i/2]+1;}for(int i=1; i<=n; i++) {stmax[0][i]=a[i];}for(int i=1; i<=log[n]; i++) {for(int j=1; j+(1<<i)-1<=n; j++) {stmax[i][j]=max(stmax[i-1][j],stmax[i-1][j+(1<<(i-1))]);}}}
int rmq_max(int l,int r) {int t=log[r-l+1];//2^log[len]>len/2;return max(stmax[t][l],stmax[t][r-(1<<t)+1]);
}
int main() {int t;scanf("%d",&t);while(t--) {scanf("%d",&n);for(int i=1; i<=n; i++) {scanf("%d",&a[i]);}init();int q;scanf("%d",&q);while(q--) {int l,r;scanf("%d%d",&l,&r);printf("%d\n",rmq_max(l,r));}}}