[HDU - 1260] Tickets (simple dp)

Tickets

Descriptions:

 

There are n individuals to buy movie tickets, if you know the time it takes to buy a ticket for each person individually, and there is still time before a person takes to buy together, at least ask how long can all be buying tickets.

Input　　　　

Give N (1 <= N <= 10), expressed set of sample N 　　　　

Is given K (1 <= K <= 2000), expressed personal buying K .. 　　　

K represents the number of the person is given a separate ticket will spend time to ensure that each number .. (0s <= Si <= 25s) 　　

It gives the number of K-1, and the man indicates that this ticket would spend time with each of the number of guaranteed .. (0s <= Si <= 50s)
the Output

For each set of data, you need to give cinema ticketing end of time, the time is 08:00:00 am start ticketing time format:. HH: MM:. SS am / pm specifically to see the sample output

Sample Input

2

2

20 25

40

1

8

Sample Output

08:00:40 am

08:00:08 am

Topic links:
https://vjudge.net/problem/HDU-1260

 

DP [i] denotes the front i need individual time a [i] denotes the i-buying their own individual time b [i] denotes the i-th individual time i-1 and the individual tickets needed

Person i he may himself buy (dp [i] == dp [i-1] + a [i]) and the front of a person or together to buy (dp [i] = dp [i-2] + b [i] )

 

AC Code

#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#define Mod 1000000007
#define eps 1e-6
#define ll long long
#define INF 0x3f3f3f3f
#define MEM(x, y) memset(x, y, sizeof(x))
#define Maxn 2000+10
using namespace std;
int n,k;
int a[Maxn];
int b[Maxn];
int dp[Maxn];
int main()
{
    cin>>n;
    while(n--)
    {
        MEM(dp,0);//初始化
        cin>>k;
        for(int i=1; i<=k; i++)
            cin>>a[i];
        for(intJ = 2 ; J <= K; J ++ ) 
            CIN >> B [J]; 
        DP [ . 1 ] = A [ . 1 ]; 
        DP [ 2 ] = min (A [ . 1 ] + A [ 2 ], B [ 2 ] );
         for ( int I = . 3 ; I <= K; I ++) // starting from the third person DP 
            DP [I] = min (DP [I- . 1 ] + A [I], DP [I- 2 ] + B [I]);
         int H DP = [K] / 3600 + . 8 ; // calculation time 
        int m = DP [K]% 3600 /60;
        int s=dp[k]%60;
        if(h>12)
            printf("%02d:%02d:%02d pm\n", h-12, m, s);
        else
            printf("%02d:%02d:%02d am\n", h, m, s);
    }
    return 0;
}