Long Path - CodeForces 407B dp

最新推荐文章于 2020-11-15 19:40:46 发布

提比-我有特殊的AC技巧

最新推荐文章于 2020-11-15 19:40:46 发布

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分类专栏： CodeForces dp动态规划和递推

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本文介绍了一道关于迷宫中寻找特定路径数量的问题。通过使用动态规划的方法，解决了小Vasya如何从迷宫的第一间房到达最后一间房的问题，并提供了一段AC代码作为解答。

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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 numberpi, where 1 ≤ pi ≤ 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 pi), 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 pi (1 ≤ pi ≤ i). Each pidenotes 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
4

input
4
1 1 2 3

output
20

input
5
1 1 1 1 1

output
62

思路：这题可以说是dp，也可以说是找规律，直接看代码吧。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long ll;
long long num[1010],mod=1000000007;
int pos[1010];
int main()
{ int T,t,n,m,i,j,k;
  scanf("%d",&n);
  for(i=1;i<=n;i++)
   scanf("%d",&pos[i]);
  num[1]=0;
  num[2]=2;
  for(i=3;i<=n+1;i++)
  { num[i]=num[i-1]*2+mod*2-num[pos[i-1]]+2;
    num[i]%=mod;
  }
  printf("%I64d\n",num[n+1]);
}