Codeofrces C. Omkar and Waterslide (模拟 / 暴力) (Global Round 10)

传送门

题意： 给出一个序列a，每次可以选择一个非递减序列将每个元素+1，试问将整个元素变成非递减需要多少次操作？

思路： 简单的模拟题，直接找到每次出现降低地方非递减序列中的min取操作数即可。

代码实现：

#include
//#define endl '\n'
#define null NULL
#define ll long long
#define int long long
#define pii pair
#define lowbit(x) (x &(-x))
#define ls(x) x<<1
#define rs(x) (x<<1+1)
#define me(ar) memset(ar, 0, sizeof ar)
#define mem(ar,num) memset(ar, num, sizeof ar)
#define rp(i, n) for(int i = 0, i < n; i ++)
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define pre(i, n, a) for(int i = n; i >= a; i --)
#define IOS ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
const int way[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
using namespace std;
const int  inf = 0x7fffffff;
const double PI = acos(-1.0);
const double eps = 1e-6;
const ll   mod = 1e9 + 7;
const int  N = 2e5 + 5;

int t, n, a[N];

signed main()
{
    IOS;

    cin >> t;
    while(t --){
        cin >>n;
        for(int i = 1; i <= n; i ++) cin >> a[i];
        int ans = 0;
        for(int i = 1; i < n; i ++){
            if(a[i] <= a[i+1]) continue;
            int minn = a[i], j = i+1;
            while(a[j-1] > a[j] && j <= n){
                minn = min(minn, a[j]);
                j ++;
            }
            ans += a[i] - minn;
            i = j-1;
        }
        cout << ans << endl;
    }

    return 0;
}