math 轉移 else clas really spa rest there ++

ickets

TimeLimit: 2000/1000 MS (Java/Others) MemoryLimit: 65536/32768 K (Java/Others) 64-bit integer IO format:%I64d 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. 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. SampleInput

2
2
20 25
40
1
8

SampleOutput

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

狀態轉移方程為dp[i]=min(dp[i-1]+s[i],dp[i-2]+d[i])
 

 
代碼如下：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include  <map>
#include <iostream>
using namespace std;
typedef long long ll;
const int MAXN=2000+10;
int si[MAXN];
 
int di[MAXN];
int dp[MAXN];
#define INF 0x3f3f3f3f
int main()
{
    int t,k,a,b,c,tmp;
    scanf("%d",&t);
    while(t--)
    {
      scanf("%d",&k);
      for(int i=1;i<=k;i++)
        scanf("%d",&si[i]);
      for(int i=1;i<=k-1;i++)
        scanf("%d",&di[i]);
      for(int i=1
 
;i<=k;i++)
        dp[i]=INF;
        dp[1]=si[1];
      for(int i=2;i<=k;i++)
          dp[i]=min(dp[i-1]+si[i],dp[i-2]+di[i-1]);
      tmp=dp[k];
      a=tmp/3600+8;
      tmp=tmp%3600;
      b=tmp/60;
      tmp=tmp%60;
      c=tmp;
      if(a<12)
        printf("%02d:%02d:%02d am\n",a,b,c);
      else
        printf("%02d:%02d:%02d pm\n",a-12,b,c);
    }
    return 0;
}

一開始一不小心寫成了 dp[i]=min(dp[i-2]+si[i-1]+si[i],dp[i-2]+di[i-1]);

導致後面debug了很久

HDU 1260 TICKETS （簡單DP)