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
很裸的区间最值查询问题，所以直接上st了。
#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));
}
}
}