hdoj--1260--Tickets(简单dp)
Tickets
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2146 Accepted Submission(s): 1039
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.
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.
2
2
20 25
40
1
8
08:00:40 am
08:00:08 am
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define MAX 101000
int dp[MAX],man[MAX],two[MAX];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n;
scanf("%d",&n);
memset(man,0,sizeof(man));
memset(two,0,sizeof(two));
for(int i=1;i<=n;i++)
scanf("%d",&man[i]);
for(int i=2;i<=n;i++)
scanf("%d",&two[i]);
memset(dp,0,sizeof(dp));
dp[1]=man[1];
for(int i=2;i<=n;i++)
dp[i]=min(dp[i-1]+man[i],dp[i-2]+two[i]);
int p=28800+dp[n];
int h=p/3600;
int m=p%3600/60;
int s=(p%3600)%60;
if(p>=43200)
{
if(h>12)
h-=12;
printf("%02d:%02d:%02d pm\n",h,m,s);
}
else
printf("%02d:%02d:%02d am\n",h,m,s);
}
return 0;
}
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