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 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 #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; }