您的当前位置：首页 Codeforces 407B Long Path【dp】好题

Codeforces 407B Long Path【dp】好题

来源：宝玛科技网

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 p_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_i 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

Input

4
1 1 2 3

Output

Input

5
1 1 1 1 1

Output

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:///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 __int
int 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);
    }
}

因篇幅问题不能全部显示，请点此查看更多更全内容

查看全文