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  • hdu5443 The Water Problem

    The Water Problem

    Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
    Total Submission(s): 602    Accepted Submission(s): 485


    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(T10) indicating the number of test cases. For each test case, there is a number n(0n1000) 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(0q1000) representing the number of queries. After that, there will be q lines with two integers l and r(1lrn) 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
     
    Source
     
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    区间求最大值
    次裸裸的RMQ,套上模板即过
    #include<stdio.h>
    #include<string.h>
    #include<iostream>
    #include<math.h>
    #include<algorithm>
    using namespace std;
    #define N 50005
    int dpmin[N][20],dpmax[N][20];
    int main()
    {    
        int t;
        scanf("%d",&t);
        while(t--){
        int i, j, n, m;
        scanf("%d",&n);
        memset(dpmin,0,sizeof(dpmin));
        memset(dpmax,0,sizeof(dpmax));
        for(i=1; i<=n; i++)
        {
            scanf("%d", &dpmin[i][0]);
            dpmax[i][0]=dpmin[i][0];
        }
        int mm=floor(log(1.0*n)/log(2.0));
        for(j=1; j<=mm; j++)
            for(i=1; i<=n; i++)
            {
                if((i+(1<<(j-1)))<=n)
                {
        //            dpmin[i][j]=min(dpmin[i][j-1], dpmin[i+(1<<(j-1))][j-1]);
                    dpmax[i][j]=max(dpmax[i][j-1], dpmax[i+(1<<(j-1))][j-1]);
                }
            }
        int x, y;
            
            scanf("%d",&m);
        for(int i=1; i<=m; i++)
        {
            scanf("%d%d", &x, &y);
            int  mid=floor(log(y*1.0-x+1)/log(2.0));
            int maxnum=max(dpmax[x][mid], dpmax[y-(1<<mid)+1][mid]);
        //    int minnum=min(dpmin[x][mid], dpmin[y-(1<<mid)+1][mid]);
            printf("%d
    ", maxnum);
        }
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/13224ACMer/p/4812142.html
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