Codeforces 957D - Riverside Curio 【思维】

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最新推荐文章于 2024-07-12 14:52:05 发布

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分类专栏：思维脑洞文章标签： Codeforces Round#472 957D-Riverside Curio

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本文介绍了一种关于河流观测的算法问题，通过记录每日河流水位变化及已标记的线上方标记数量，来求解最小化的线下标记总数。采用逆向思维与累积更新策略，实现最优解。

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D. Riverside Curio

time limit per test 1second

memory limit per test 256megabytes

Arkady decides to observe a river for n consecutivedays. The river's water level on each day is equal to some real value.

Arkady goes to the riverside each dayand makes a mark on the side of the channel at the height of the water level,but if it coincides with a mark made before, no new mark is created. The waterdoes not wash the marks away. Arkady writes down the number of marks strictlyabove the water level each day, on the i-th day this value is equal to m_i.

Define d_i as the number of marks strictly underthe water level on the i-th day. You are to find out the minimum possible sum of d_i over all days.There are no marks on the channel before the first day.

Input

The first line contains a singlepositive integer n (1 ≤ n ≤ 10⁵) — thenumber of days.

The second line contains n space-separatedintegers m₁, m₂, ..., m_n (0 ≤ m_i < i) — the number of marks strictly above the water on each day.

Output

Output one single integer — the minimum possible sum of the number of marks strictly below the water level among all days.

Examples

Input

Copy

6
0 1 0 3 0 2

Output

Input

5
0 1 2 1 2

Output

Input

5
0 1 1 2 2

Output

Note

In the first example, the following figure shows an optimal case.

Note that on day 3, a new mark should be created because if not, there cannot be 3 marks above water on day 4. The totalnumber of marks underwater is 0 + 0 + 2 + 0 + 3 + 1 = 6.

In the second example, the following figure shows an optimal case.

【题意】

有一个人每天会去观察一条河流的水位，每次在当前水位上画一条线(如果重复则不画)

现在告诉你他每天去看时，当前水位的上方(严格大于)画了几条线。

问每天去看时，当前水位的下方(严格小于)画线条数d[i]总和的最小值。

【思路】

我们用sum[i]表示，前i天一共画了多少条线，由于sum[i]=d[i]+a[i]+1,那么最小化sigma(d[i])就转化为了最小化sigma(sum[i)。

于是我们尽量使得每个sum[i]小，由于sum[i]必须满足：

sum[i]>=max(sum[i-1],a[i]+1)
sum[i]>=sum[i+1]-1 每天最多画一条线

条件1很好满足，为了满足条件2,我们需要从最后一天开始往上更新即可

然后再正向更新一下，保证sum[i]>=sum[i-1]

最后代入公式即可。

#include <cstdio>
#include <cmath>
#include <queue>
#include <cstring>
#include <algorithm>
using namespace std;
#define mst(a,b) memset((a),(b),sizeof(a))
#define rush() int T;scanf("%d",&T);while(T--)

typedef long long ll;
const int maxn = 100005;
const ll mod = 1e9+7;
const ll INF = 1e18;
const double eps = 1e-6;

int n;
ll a[maxn];
ll sum[maxn];

int main()
{
    scanf("%d",&n);
    for(int i=0;i<n;i++) scanf("%lld",&a[i]);
    ll now=0;
    for(int i=n-1;i>=0;i--)
    {
        now=max(now-1,a[i]+1);
        sum[i]=now;
    }
    ll ans=sum[0]-a[0]-1;
    for(int i=1;i<n;i++)
    {
        sum[i]=max(sum[i],sum[i-1]);
        ans+=(sum[i]-(a[i]+1));
    }
    printf("%lld\n",ans);
}