摘要：
Tickets 如果是这样的话，我会感谢你的帮助。输入有两种不同的场景，每个场景由三行组成：1）一个整数K表示总人数；2） 亲属号码表示每个人花费的时间；3） （K-1）整数表示两个相邻（相似）人一起购买两张票所需的时间。输出对于每种情况，请告诉Joeathati他们可能会尽快回家。每天早上8点开始上班。格式HH:MM:SSam|pm.SampleInput2220254018SampleOutput08:00:40am08:00:08amSource浙江理工大学第四届本科生课程设计竞赛推荐JGMachining |我们为您精心挑选了几个类似的问题：11601231107410691159我觉得dp真的很强大，dp[I]=min。在整个周期中比较最佳解决方案。Dp[i-1]+num1[i]是每次未买票的人花费的时间。dp[i-2]+num2[i-1]是两个人一起买票的时间（两次）--------据说中午是下午，这是有争议的。
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1941 Accepted Submission(s): 938
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.
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(相近的)people to buy two tickets together.
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(相近的)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
Source
Recommend
感觉dp真强大, dp[i]=min(dp[i-1]+num1[i] , dp[i-2]+num2[i-1]). 通过循环来比较求的最优解。 dp[i-1]+num1[i]每次+一个未买票的人所耗时间， dp[i-2]+num2[i-1]为两人共同买票（两次）所用时间。 求最小值。
-------- 正午据说是pm , 说法存在争议。
1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #define min(a, b) a<b?a:b 5 using namespace std; 6 int main() 7 { 8 int t; 9 int num1[2020], num2[2020], dp[2020]; 10 scanf("%d", &t); 11 while(t--) 12 { 13 int m; 14 scanf("%d", &m); 15 for(int i = 1; i <= m; i++) 16 scanf("%d", &num1[i]); 17 for(int i = 1; i <= m-1; i++) 18 scanf("%d", &num2[i]); 19 memset(dp, 0, sizeof(dp)); 20 dp[1] = num1[1]; 21 for(int i = 2; i <= m; i++) 22 dp[i] = min(dp[i-1] + num1[i], dp[i-2] + num2[i-1]); 23 int b, c, d; 24 b = dp[m] % 60; 25 c = dp[m] / 60 % 60; 26 d = dp[m] /3600; 27 d += 8; 28 char c1, c2; 29 if(d < 12) 30 { 31 c1 = 'a'; 32 c2 = 'm'; 33 } 34 else 35 { 36 d %= 12; 37 c1 = 'p'; 38 c2 = 'm'; 39 } 40 printf("%02d:%02d:%02d %c%c ", d, c, b, c1, c2); 41 } 42 return 0; 43 }