HDU 1260 Tickets

HDU 1260 Tickets

Time limit 1000 ms
Memory limit 32768 kB
OS Windows

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Problem Description

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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

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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

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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

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2
2
20 25
40
1
8

Sample Output

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08:00:40 am
08:00:08 am

Submit

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#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;#define maxn 2010#define INF 0x3f3f3f3fint n[maxn], a[maxn];int dp[maxn];int main(){    int T;    int h, m, s;    scanf("%d", &T);    while(T--)    {        int N;        scanf("%d", &N);        dp[0] = 0;        for(int i = 1; i <= N; i++)        {            scanf("%d", &a[i]);        }        memset(n, INF, sizeof(n));        for(int i = 2; i <= N; i++)            scanf("%d", &n[i]);        dp[1] = a[1];        for(int i = 2; i <= N; i++)            dp[i] = min(dp[i-1]+a[i], dp[i-2]+n[i]);//一次买1张或1次买2张        s = dp[N];        h = s / 3600;        s %= 3600;        m = s / 60;        s %= 60;        h += 8;        h %= 24;        if(h <= 12)            printf("%02d:%02d:%02d am\n", h, m, s);        else            printf("%02d:%02d:%02d pm\n", h, m, s);    }    return 0;}