Codeforces Round #239 (Div. 1)

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原文链接：http://www.cnblogs.com/lahblogs/p/3635743.html

本文介绍了一个迷宫问题的动态规划解决方案。问题设定为Vasya在一个由多个房间组成的迷宫中，需要从第一个房间走到最后一个房间。每个房间有两个门，选择门的规则依据已标记次数的奇偶性决定。文章提供了完整的代码实现，包括输入输出和状态转移方程。

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One day, little Vasya found himself in a maze consisting of (n + 1) rooms, numbered from 1 to (n + 1). Initially, Vasya is at the first room and to get out of the maze, he needs to get to the (n + 1)-th one.

The maze is organized as follows. Each room of the maze has two one-way portals. Let's consider room number i (1 ≤ i ≤ n), someone can use the first portal to move from it to room number (i + 1), also someone can use the second portal to move from it to room numberp_i, where 1 ≤ p_i ≤ i.

In order not to get lost, Vasya decided to act as follows.

Each time Vasya enters some room, he paints a cross on its ceiling. Initially, Vasya paints a cross at the ceiling of room 1.
Let's assume that Vasya is in room i and has already painted a cross on its ceiling. Then, if the ceiling now contains an odd number of crosses, Vasya uses the second portal (it leads to room p_i), otherwise Vasya uses the first portal.

Help Vasya determine the number of times he needs to use portals to get to room (n + 1) in the end.

Input

The first line contains integer n (1 ≤ n ≤ 10³) — the number of rooms. The second line contains n integers p_i (1 ≤ p_i ≤ i). Each p_idenotes the number of the room, that someone can reach, if he will use the second portal in the i-th room.

Output

Print a single number — the number of portal moves the boy needs to go out of the maze. As the number can be rather large, print it modulo 1000000007 (10⁹ + 7).

Sample test(s)

input

2
1 2

output

input

4
1 1 2 3

output

input

5
1 1 1 1 1

output

题意：Vasya在一个由各种房间组成的迷宫里，房间的编号为（1~n），开始Vasya在1号房间，每个房间都有两个门，第一扇门通往下一个房间，第二扇门通往任意一个房间，Vasya决定每到一个房间就在该房间标记一次，如果标记数为奇数就打开第二扇门，否则打开第一扇门，问Vasya走出迷宫需要的步数。

思路：dp[i]表示偶数次到第i个房间的步数，则状态转移方程为dp[i] = dp[i-1]+1+dp[i-1]-dp[a[i]-1]+1;

 1 #include <stdio.h>
 2 #include <iostream>
 3 #include <string.h>
 4 using namespace std;
 5 const int N=1002;
 6 const int MOD=1000000007;
 7 long long dp[N],a[N];
 8 int main()
 9 {
10     int n;
11     while(~scanf("%d",&n))
12     {
13         for (int i = 1; i <= n; i++)
14             cin>>a[i];
15         dp[0] = -1;
16         dp[1] = 1;
17         for (int i = 2; i <= n; i++)
18         {
19             dp[i] = (dp[i-1]+1+dp[i-1]-dp[a[i]-1]+1)%MOD;
20         }
21         cout<<((dp[n]+1)%MOD+MOD)%MOD<<endl;
22     }
23     return 0;
24 }