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. 

08:00:40 am

08:00:08 am

大意：
　　 卖电影票，先输入t，代表t组测试数据，没组测试数据，第一行输入一个n代表n个人，第二行为n个整数代表票卖给每个人用的时间，第二行为n-1个数，代表卖给两个人（第i个人和第i+1个人，与上面的人对应）　　 用的时间；
　　 例如：
　　　　　　3
　　　　　　a[1] a[2] a[3]
　　　　　　b[1] b[2]
　　　a[i]代表卖票给第i个人用的时间，b[i]代表同时卖票给第i个人和第i+1个人用的时间。
思路：
　　定义数组a[i]记录单个人买票的时间b[i]记录两个人买票的时间，d[i]记录i个人买票的最短时间，根据
　　d[i]=min(d[i-2]+b[i-1],d[i-1]+a[i])（i-2个人用的最短时间加第i-1和第i个人一起买票与i-1个人用的最短时间加第i个人买票的时间），   求出n个人买票用的最短时间，最后控制输出格式。

 #include<cstdio>

 #include<algorithm>

 using namespace std;

 int main()

 {

     int t;

     scanf("%d",&t);

     while(t--)

     {

         int a[],b[],d[];

         int n;

         scanf("%d",&n);

         for(int i =  ; i <= n ; i++)

         {

             scanf("%d",&a[i]);

         }

         for(int i =  ; i <= n- ; i++)

         {

             scanf("%d",&b[i]);

         }

         d[]=;

         d[]=a[];

         for(int i =  ; i <= n ; i++)

         {

             d[i]=min(d[i-]+b[i-],d[i-]+a[i]);        //表示i个人买票用的最短时间

         }

         int s=d[n]%;

         int m=d[n]/%;

         int t=+d[n]//;

         printf("%02d:%02d:%02d am\n",t,m,s);

     }

 }