hduoj 5443 The Water Problem【线段树】

The Water Problem

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1903    Accepted Submission(s): 1508

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 witha1,a2,a3,...,an representing the size of the water source. Given a set of queries each containing2 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

3110011 151 2 3 4 551 21 32 43 43 531 999999 141 11 22 33 3

 

Sample Output

1002344519999999999991
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int N = 1000 + 10;int maxn[N][30];int a[N];int num,q;void RMQ(){for(int i = 1; i <= num; i++)   maxn[i][0] = a[i];for(int j = 1; (1 << j) <= num; j++)   for(int i = 1; (i + (1 << j) - 1) <= num; i++){     maxn[i][j] = max(maxn[i][j-1],maxn[i+(1<<(j-1))][j-1]);   }}int Query(int L, int R){int k = 0;while((1 << (k + 1)) <= R - L + 1)   k++;return max(maxn[L][k],maxn[R - (1 << k) + 1][k]);}int main(){int t;scanf("%d",&t);while(t--){scanf("%d",&num);for(int i = 1; i <= num; i++)   scanf("%d",&a[i]);RMQ();int l,r;scanf("%d",&q);while(q--){scanf("%d%d",&l,&r);printf("%d\n",Query(l,r));}}return 0;}