題目連結

題意

一共有n件商店，每個商店買賣東西的價格和不同，白兔想用差價掙錢，每次只能攜帶一種物品，問白兔最多賺多少的錢，和對應的交易次數

AC

兩個變數分別記錄當前買東西和買東西的剩餘財富，財富越大可以使得買下一件物品的損失更低，賣下一件東西的收益更高。列舉n個物品，維護最大財富值。

#include <bits/stdc++.h>
#define N 100005
#define P pair<int,int>
#define mk(a, b) make_pair(a, b)
#define mem(a, b) memset(a, b, sizeof(a))
 

#define pb(a) push_back(a)
#define lowbit(a) a&(-a)
#define ll long long
using namespace std;
int inf = 0x3f3f3f3f;
int mod = 1e9 + 7;
int main(){
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
#endif
    int t;
    scanf("%d", &t);
    while (t--) {
        int n;
        scanf("%d"
 
, &n);
        vector<ll> a(n + 1);
        for (int i = 1; i <= n; ++i) {
            scanf("%lld", &a[i]);
        }
        // mon_sell 記錄如果賣這件物品的最大錢數
        // mon_buy  記錄如果買這件物品的最大錢數
        ll mon_sell = 0, mon_buy = -a[1];
        // time_buy 買這件物品交換的次數
        // time_sell 賣這件物品交換的次數
 

        ll time_sell = 0, time_buy = 1;
        // 列舉每個物品
        for (int i = 2; i <= n; ++i) {
            ll selled = mon_buy + a[i];
            ll buyed = mon_sell - a[i];
            // 更新
            if (selled > mon_sell) {
                mon_sell = selled;
                time_sell = time_buy + 1;
            }else if(selled == mon_sell) {
                time_sell = min(time_sell, time_buy + 1);
            }

            // 更新
            if (buyed > mon_buy) {
                mon_buy = buyed;
                time_buy = time_sell + 1;
            }else if(buyed == mon_buy) {
                time_buy = min(time_buy, time_sell + 1);
            }
        }
        printf("%lld %lld\n", mon_sell, time_sell);  
    }
    return 0;
}