AcWing 4553. 卖票

设 \(f_i\) 表示对于前 \(i\) 个顾客开票所需的最短时间，\(f_i=\max\{f_{i-1}+a_i,f_{i-2}+b_i\}\)

左边表示单选，右边表示一块选。

// #define FILE_INPUT
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

#define rep(i, a, b) for (int i = a, END##i = b; i <= END##i; i++)
#define per(i, a, b) for (int i = a, END##i = b; i >= END##i; i--)

void Init();
void Solve();

signed main() {
    // cin.sync_with_stdio(0);
    // cin.tie(0), cout.tie(0);

    #ifdef FILE_INPUT
        freopen("input.in", "r", stdin);
    #endif

    int T = 1;
    cin >> T;
    while (T--) {
        Init();
        Solve();
    }
    return 0;
}

using LL = long long;
using ULL = unsigned long long;

const int Mod = 1e9 + 7;
const int Inf = 0x3f3f3f3f;
const LL InfLL = 0x3f3f3f3f3f3f3f3f;

const int N = 1e6 + 10;
int f[N], n, a[N], b[N];

void Init() {
}

void Solve() {
    cin >> n;
    rep(i, 1, n) cin >> a[i];
    rep(i, 2, n) cin >> b[i];
    f[1] = a[1];
    rep(i, 2, n) f[i] = min(f[i - 2] + b[i], f[i - 1] + a[i]);
    int s = f[n];
    int mi = s / 60; s %= 60;
    int h = mi / 60 + 8; mi %= 60;
    if (h < 12) printf("%02d:%02d:%02d am\n", h, mi, s);
    else if (h == 12) printf("%02d:%02d:%02d pm\n", h, mi, s);
    else printf("%02d:%02d:%02d pm\n", h - 12, mi, s);
}