zoukankan      html  css  js  c++  java
  • HDU Problem 1260 Tickets 【dp】

    Tickets

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 3176    Accepted Submission(s): 1564

    Problem Description
    Jesus, what a great movie! Thousands of people are rushing to the cinema. However, this is really a tuff time for Joe who sells the film tickets. He is wandering when could he go back home as early as possible.
    A good approach, reducing the total time of tickets selling, is let adjacent people buy tickets together. As the restriction of the Ticket Seller Machine, Joe can sell a single ticket or two adjacent tickets at a time.
    Since you are the great JESUS, you know exactly how much time needed for every person to buy a single ticket or two tickets for him/her. Could you so kind to tell poor Joe at what time could he go back home as early as possible? If so, I guess Joe would full of appreciation for your help.
     
    Input
    There are N(1<=N<=10) different scenarios, each scenario consists of 3 lines:
    1) An integer K(1<=K<=2000) representing the total number of people;
    2) K integer numbers(0s<=Si<=25s) representing the time consumed to buy a ticket for each person;
    3) (K-1) integer numbers(0s<=Di<=50s) representing the time needed for two adjacent people to buy two tickets together.
     
    Output
    For every scenario, please tell Joe at what time could he go back home as early as possible. Every day Joe started his work at 08:00:00 am. The format of time is HH:MM:SS am|pm.
     
    Sample Input
    2 2 20 25 40 1 8
     
    Sample Output
    08:00:40 am 08:00:08 am
     
    #include <bits/stdc++.h>
    #define MAXN 100005
    using namespace std;
    const int INF = 1e9;
    int dp[MAXN], ar1[MAXN], ar2[MAXN];
    int n;
    int main() {
        int t; scanf("%d", &t);
        while (t--) {
            scanf("%d", &n);
            for (int i = 1;i <= n; i++)
                scanf("%d", &ar1[i]);
            for (int i = 1;i < n; i++)
                scanf("%d", &ar2[i]);
            dp[1] = ar1[1]; dp[0] = 0;
            for (int i = 2; i <= n; i++) {
                dp[i] = min(dp[i-1] + ar1[i], dp[i-2]+ar2[i-1]);
            }
            int h, m, s;
            s = dp[n]%60; m = dp[n]/60;
            h = m/60; m %= 60; h += 8;
            int ma = 0;
            while (h > 12) {
                ma++; h -= 12;
            }
            printf("%02d:%02d:%02d ", h, m, s);
            if (ma%2 == 0)  printf("am
    ");
            else printf("pm
    ");
        }
        return 0;
    }

  • 相关阅读:
    GitHub 如何创建 Access Token
    Fact Table 数据表什么意思
    高基数数据特性是什么意思
    Edge 浏览器的隐藏 URL QR 生成器
    Apache Druid 安装的时候进行 Java 版本校验没有输出
    Apache Druid 简介
    如何在 Discourse 中批量移动主题到不同的分类中
    素材
    Drawable转bitmap
    Drawable与 Bitmap 转换总结
  • 原文地址:https://www.cnblogs.com/cniwoq/p/6770849.html
Copyright © 2011-2022 走看看