D. Riverside Curio
time limit per test 1 second
memory limit per test 256 megabytes
input standard input
output standard output
Arkady decides to observe a river for n consecutive days. The river’s water level on each day is equal to some real value.
Arkady goes to the riverside each day and 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 water does not wash the marks away. Arkady writes down the number of marks strictly above the water level each day, on the i-th day this value is equal to mi.
Define di as the number of marks strictly under the water level on the i-th day. You are to find out the minimum possible sum of di over all days. There are no marks on the channel before the first day.
Input
The first line contains a single positive integer n (1 ≤ n ≤ 105) — the number of days.
The second line contains n space-separated integers m1, m2, …, mn (0 ≤ mi < 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
6
0 1 0 3 0 2
output
6
input
5
0 1 2 1 2
output
1
input
5
0 1 1 2 2
output
0
题意：你用n天时间去观察一条河，这条河的水位每天都有可能改变，所有你每天都会在当前水位上画一条线来标注（这个线永远都不会消失），当然啦，如果这个水位刚好和之前某天水位一致，也就是已经有线和水面平齐，那么当然就不需要再标记了。现在告诉你每天可以看到的水位线数量m[i]，求出∑d[i]的最小值，其中d[i]表示第i天水底下的标记线的个数
解析:
设c[i]表示第i次到河边后总共画了多少条线，那么有c[i] = m[i]+d[i]+1
由题意c[i]必须非严格递增，并且相邻两个数差距不超过1
方法很简单，只要先把它变成非严格递增，如果m[i]< m[i-1]，就让m[i] = m[i-1]，然后再从后往前遍历如果m[i]+1< m[i+1]，那么m[i] = m[i+1]-1满足第二个条件就行了
#include<bits/stdc++.h>
#define ll long long
#define inf 0x3f3f3f3f
#define pb push_back
#define mod 1000000007
#define rep(i,a,b) for(int i=a;i<b;i++)
#define rep1(i,b,a) for(int i=b;i>=a;i--)
using namespace std;
const int N=1e5+100;
int a[N];
int main()
{
ios::sync_with_stdio(false);
ll ans;
int n,i;
cin>>n;
for(i=1;i<=n;i++)
cin>>a[i];
ans = 0;
for(i=1;i<=n;i++)
{
if(a[i]<a[i-1])
{
ans+=a[i-1]-a[i];
a[i]=a[i-1];
}
}
for(i=n-1;i>=1;i--)
{
if(a[i]+1<a[i+1])
{
ans+=a[i+1]-a[i]-1;
a[i]=a[i+1]-1;
}
}
cout<<ans<<endl;
return 0;
}