牛客网暑期ACM多校训练营（第二场） D money 思维

链接：https://www.nowcoder.com/acm/contest/140/D
来源：牛客网

White Cloud has built n stores numbered from 1 to n.
White Rabbit wants to visit these stores in the order from 1 to n.
The store numbered i has a price a[i] representing that White Rabbit can spend a[i] dollars to buy a product or sell a product to get a[i] dollars when it is in the i-th store.
The product is too heavy so that White Rabbit can only take one product at the same time.
White Rabbit wants to know the maximum profit after visiting all stores.
Also, White Rabbit wants to know the minimum number of transactions while geting the maximum profit.
Notice that White Rabbit has infinite money initially.

输入描述:

The first line contains an integer T(0<T<=5), denoting the number of test cases.
In each test case, there is one integer n(0<n<=100000) in the first line,denoting the number of stores.
For the next line, There are n integers in range [0,2147483648), denoting a[1..n].

输出描述:

For each test case, print a single line containing 2 integers, denoting the maximum profit and the minimum number of transactions.示例1

输入

复制

1
5
9 10 7 6 8

输出

复制

3 4
 
分析：我每次只能每一个，然后必须卖掉才能买下一个，中间的利润要最大，交易次数最小
我们可以在坐标系里标出每个点，横坐标点的位置，纵坐标点的值
然后我们发现我们所要求得就是每条非递减线段的值
如下图中，我们要求的就是AB+DE

注意点值相同的时候要加上这条非递减数列的最后的值，如果不判相等中间断了结果就错了

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
const int maxn = 1e5 + 10;
const int mod = 10000007;
typedef long long ll;
int t,n;
ll a[maxn],ans,s;
int main() {
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    cin>>t;
    while(t--){
        ans=s=0;
        cin>>n;
        for(int i=1;i<=n;i++) cin>>a[i];
        if(n<2) {
            cout<<"0 0"<<endl;
            continue;
        }
        int fg=0;
        for(int i=2;i<=n;i++) {
            if(a[i]==a[i-1]) continue;
            if(a[i]>=a[i-1]) ans+=(a[i]-a[i-1]),fg=1;
            else if(fg) s+=2,fg=0;
        }
        if(a[n]>=a[n-1]&&fg) s+=2;
        cout<<ans<<" "<<s<<endl;
    }
    return 0;
}