Codeforces 407B Long Path【dp】好题

B. Long Path

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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 number
pi, 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
pi denotes 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).

Examples

Input

2
1 2

Output

4

Input

41 1 2 3

Output

20

Input

5
1 1 1 1 1

Output

62

题目大意：

起点为1，终点为n+1，对应第i个各点，如果我奇数次到达，那么下一步走到a【i】的位子，如果是偶数次到达，那么下一步走到a【i】+1的位子。

问从1走到n+1一共需要走多少步

思路：

思路来自巨牛：http://blog.csdn.net/qq_24451605/article/details/48465609

1、设定dp【i】【j】表示从i走到j一共需要多少步。

2、考虑到a【i】<=i，那么要想到i+1这个格点去，那么一定是从第i个格点走过去的。

那么dp【i】【j】=dp【i】【j-1】+dp【a【j-1】】【j-1】+1；



dp【i】【j-1】表示从i到j-1需要的步数，同时也是第一次到达j-1这个位子的步数，那么我们想要从j-1到达j，那么我们明显需要从a【j-1】再走到j-1之后，就是第二次到达了j-1这个位子，那么接下来就可以走到j了。

3、记忆化搜索来实现代码即可。

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
#define ll __int64int a[10000];
ll dp[1005][1005];
int Dfs(int u,int v)
{
if(u==v)
{
dp[u][v]=1;
return dp[u][v];
}
if(dp[u][v]==-1)
{
dp[u][v]=0;
dp[u][v]=Dfs(u,v-1)+Dfs(a[v-1],v-1)+1;
dp[u][v]%=1000000000+7;
return dp[u][v];
}
else return dp[u][v];
}
int main()
{
int n;
while(~scanf("%d",&n))
{
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
memset(dp,-1,sizeof(dp));
Dfs(1,n+1);
printf("%d\n",dp[1][n+1]-1);
}
}