5 seconds
256 megabytes
standard input
standard output
Sonya was unable to think of a story for this problem, so here comes the formal description.
You are given the array containing n positive integers. At one turn you can pick any element and increase or decrease it by 1. The goal is the make the array strictly increasing by making the minimum possible number of operations. You are allowed to change elements in any way, they can become negative or equal to 0.
The first line of the input contains a single integer n (1 ≤ n ≤ 3000) — the length of the array.
Next line contains n integer ai (1 ≤ ai ≤ 109).
Print the minimum number of operation required to make the array strictly increasing.
Note
In the first sample, the array is going to look as follows:
2 3 5 6 7 9 11
|2 - 2| + |1 - 3| + |5 - 5| + |11 - 6| + |5 - 7| + |9 - 9| + |11 - 11| = 9
And for the second sample:
1 2 3 4 5
|5 - 1| + |4 - 2| + |3 - 3| + |2 - 4| + |1 - 5| = 12
题意：
思路：
而本题的严格如何转化非严格，直接加一行代码，a[i]-=i；
原理
推导过程大致如下：a[i] < a[i+1] —> a[i] <= a[i+1]-1 —> a[i]-i <= a[i+1]-(i+1)—> b[i]<=b[i+1]
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
#define pb push_back
typedef long long ll;
using namespace std;
const int N = 3e3+;
const int M = ;
int n,a[N],b[N];
ll dp[N][N]; int main(){
scanf("%d",&n);
for(int i=; i<=n; i++){
scanf("%d",&a[i]);
a[i]-=i;
b[i]=a[i];
}
sort(b+, b+n+);
for(int i=; i<=n; i++){
ll minn=dp[i-][];
for(int j=; j<=n; j++){
minn=min(minn, dp[i-][j]);
dp[i][j]=abs(a[i]-b[j])+minn;
}
}
ll ans=dp[n][];
for(int i=; i<=n; i++){
ans=min(ans, dp[n][i]);
}
printf("%lld\n",ans);
return ;
}