[CSP-S Simulation Test]: Olga not stop head (DP)

Topic Portal (internal title 125)

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Input Format

　　A first line integer $ n $, in the meaning of the same title.
　　The second line n-$ $ integer, denotes the number of $ I $ $ $ P_i, the meaning of the same title.

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Output Format

　　A line integer, $ represents the result modulo the answer to $ 1,000,000,007 (10 ^ 7 + 9).

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Sample

Sample input 1:

2
1 2

Sample output 1:

4

Sample input 2:

4
1 1 2 3

Sample Output 2:

20

Sample input 3:

5
1 1 1 1 1

Sample output 3:

62

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Data range and tips

Sample $ 1 $ explanation:

　　Olga in the initial position $ 1 $, because this is his first time to reach $ 1 $ 1 $ $ position, so the first step he would come to $ p_i = $ 1, $ 1 $ position at this time has reached twice, so the second step Olga will come to $ 1 + 1 = 2 $ position.
　　Similarly, Olga will go next $ p_2 = 2 $, $ 2 + 1 = 3 $, total cost $ 4 $ Step $ n + 1 $ reaches position.

Sample $ 3 $ explanation:

　　Do you really have the heart to let me list them all right, say so, do not stop ah (refer lazy).

data range:

　　For $ 10 \% $ data, to ensure that $ 1 \ leqslant n \ leqslant 20 $;
　　for additional $ 10 \% $ data satisfies $ p_i = i $;
　　for data additional $ 20 \% $ meet $ p_i = 1 $;
　　Further data for $ 20 \% $ ensure $ 1 \ leqslant n \ leqslant 1,000 $;
　　for $ 100 \ $% of the data, to ensure that $ 1 \ leqslant n \ leqslant 1,000,000,1 \ leqslant p_i \ leqslant i $.

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answer

Consider $ DP $, defined $ dp [i] $ $ 1 represents the first $ $ I $ to the number of steps.

Then we can list the state transition equation:

$$dp[i+1]=2\times dp[i]-dp[p[i]]+2$$

因为$p[i]\leqslant i$，所以直接递推即可。

时间复杂度：$\Theta(n)$。

期望得分：$100$分。

实际得分：$100$分。

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代码时刻

#include<bits/stdc++.h>
using namespace std;
const int mod=1000000007;
int n;
long long dp[1000002];
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{int x;scanf("%d",&x);dp[i+1]=(2*dp[i]-dp[x]+2+mod)%mod;}
	printf("%d",dp[n+1]);
	return 0;
}

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rp++