Tickets 1260

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

题意：就是一个人卖票，他8点开始上班，票既可以单卖也可以一次卖两张。。。他最快可以几点下班。（第三行是单卖个第n个人的时间，第四行是一起卖给第n个人和第n+1人的时间）

思路：首先要注意输入一组数据的时候，只需要输入一个值，用dp[ ]储存最小和，然后递推。。。（最后别忘记输出am和pm，就是因为这个地方WA了一次）

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int main(){int n,m,d,i,k,e,f,a[2010],b[2010],dp[2010];scanf("%d",&n);while(n--){memset(dp, 0, sizeof(dp));          memset(a, 0, sizeof(a));          memset(b, 0, sizeof(b));scanf("%d",&m);if(m==1){scanf("%d",&dp[1]);}else{for(i=1;i<=m;i++){scanf("%d",&a[i]);}for(i=2;i<=m;i++){scanf("%d",&b[i]);}dp[0]=0;dp[1]=a[1];for(i=2;i<=m;i++){dp[i]=min(dp[i-1]+a[i],dp[i-2]+b[i]);}}d=dp[m]/3600+8;e=dp[m]/60%60;f=dp[m]%60;if(d>24) d=d/24;printf("%02d:%02d:%02d ",d,e,f);if(d<12)printf("am\n");else printf("pm\n");}return 0;}