zoukankan      html  css  js  c++  java
  • NEFU 698 Post office 大概是dp?

    Post office

    Time Limit 1000ms

    Memory Limit 65536K

    description

    There are N(N<=1000) villages along a straight road, numbered from 1 to N for simplicity. We know exactly the position of every one (noted pos[i],pos[i] is positive integer and pos[i]<=10^8). The local authority wants to build a post office for the people living in the range i to j(inclusive). He wants to make the sum of |pos[k]-position_of_postoffice| (i<=k<=j) is minimum. 
    							

    input

      For each test case, the first line is n. Then n integer, representing the position of every village and in acending order. Then a integer q (q<=200000), representing the queries. Following q lines, every line consists of two integers i and j. the input file is end with EOF. Total number of test case is no more than 10.
      Be careful, the position of two villages may be the same.
    							

    output

      For every query of each test case, you tell the minimum sum.
    							

    sample_input

    3
    1 2 3
    2
    1 3
    2 3
    0
    							

    sample_output

    2
    1
    							

    hint

      Huge input,"scanf" is recommend.
    ------------------

    可以用dp进行一下预处理=。=

    但是!!!!暴力也能过!!!!!

    ------------------

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    
    using namespace std;
    
    long long a[1111];
    long long f[1111];
    long long g[1111];
    int n;
    int q;
    
    
    int main()
    {
        while (~scanf("%d",&n))
        {
            if (n==0) break;
            memset(a,0,sizeof(a));
            memset(f,0,sizeof(f));
            memset(g,0,sizeof(g));
    
            for (int i=1;i<=n;i++)
            {
                scanf("%I64d",&a[i]);
                f[i]=f[i-1]+a[i];
            }
            for (int i=n;i>=1;i--)
            {
                g[i]=g[i+1]+a[i];
            }
            scanf("%d",&q);
            while (q--)
            {
                int x,y;
                scanf("%d%d",&x,&y);
                int mid=(x+y)/2;
                long long ans=(a[mid]*(mid-x)-(f[mid-1]-f[x-1]))+((g[mid+1]-g[y+1])-a[mid]*(y-mid));
                printf("%I64d\n",ans);
            }
        }
        return 0;
    }
    





  • 相关阅读:
    树莓派3 之 启动 和 系统配置
    树莓派3 之 初次使用
    Python 资源大全中文版
    乔布斯:遗失的访谈
    CSS3j背景渐变,字体颜色渐变,以及兼容IE写法
    系统设计相关
    JSON格式要求
    VUE解决空格和空行报错的问题
    css3实现悬停波浪效果
    css3实现匀速无限滚动效果
  • 原文地址:https://www.cnblogs.com/cyendra/p/3226385.html
Copyright © 2011-2022 走看看