Tickets

Tickets

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

注意am/pm 还有时分秒的转换
思路：每到第a[i]个选择加上第a[i]个一个人和加上第a[i]和a[i-1]两个人一起的时间

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int ans[100001];
int a[100001],b[100001];
int main(){
	int t;
	scanf("%d",&t);
	while(t--){
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		memset(ans,0,sizeof(ans));
		int n;
		scanf("%d",&n);
		
		for(int i=0;i<n;i++){
			scanf("%d",&a[i]);
		}
		for(int i=0;i<(n-1);i++){
			scanf("%d",&b[i]);
		}
		
		int i=0;
		ans[0]=a[0];
		ans[1]=min(a[0]+a[1],b[0]);
		for(int i=2;i<n;i++){
			ans[i]=min(ans[i-1]+a[i],ans[i-2]+b[i-1]);
		}
		//cout<<ans[n-1]<<endl;
		int shi=8;
		int fen=0;
		int miao=0;
			shi+=(ans[n-1]/3600);
			fen+=((ans[n-1]%3600)/60);
			miao+=(ans[n-1]%60);
		if(shi<=12){
			printf("%02d:%02d:%02d am\n",shi,fen,miao);
		}
		else{
			printf("%02d:%02d:%02d pm\n",shi,fen,miao);
		}
	}
	
	return 0;
}

/*

2
10
1120 1125 1120 1145 1120 1125 1120 1145 1120 1125 
1140 1155 5511 1140 1155 1155 4110 1155 1155 
1
8

*/