hdu 5443 ST表 简单求最大最小值

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

求 l r的最大值

#include <bits/stdc++.h>using namespace std;int a[1100];int mx[1100][30];int n;void init(){    memset(mx,0,sizeof(mx));    for(int i=1;i<=n;i++)        mx[i][0]=a[i];    int k=floor(log((double)n)/log(2.0));    for(int i=1;i<=k;i++)    {        for(int j=n;j>=1;j--)        {            if(j+(1<<(i-1))<=n)            {                mx[j][i]=max(mx[j][i-1],mx[j+(1<<(i-1))][i-1]);            }        }    }}int rmq(int i,int j){    int k=floor(log((double)(j-i+1))/log(2.0));    int maxx=max(mx[i][k],mx[j-(1<<k)+1][k]);    return maxx;}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 m;        scanf("%d",&m);        while(m--)        {            int c,d;            scanf("%d%d",&c,&d);            printf("%d\n",rmq(c,d) );        }    }}