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？
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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。
【Codeofrces C. Omkar and Waterslide (模拟 / 暴力) (Global Round 10)】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 intinf = 0x7fffffff;
const double PI = acos(-1.0);
const double eps = 1e-6;
const llmod = 1e9 + 7;
const intN = 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;
}
