Codeforces 408D Long Path (DP)

题目：

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

21 2

output

input

41 1 2 3

output

input

51 1 1 1 1

output

题意：如果当前到达点i的进入次数cnt为奇数 则可以到达p[i]位置 如果是偶数 可以到达i+1位置 求从1走到n+1需要的总步数 

思路：第一次走进i房间的进入次数cnt为1 是奇数 会往后退回p[i]位置 但是此刻p[i]位置的cnt也变成了奇数 因为当时只有为偶数才能向前走 那么会退回到p[p[i]]位置 不断递归因此 dp[i]记为走到当前位置需要的总步数 状态转移方程为dp[i+1]=dp[i]+1+(dp[i]-dp[p[i]])+1 即为dp[i+1]=2*dp[i]-dp[p[i]]+2

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int inf=0x3f3f3f3f;
const int maxn=1e3+10;
const int mod=1e9+7;
int n;
int p[maxn];
ll dp[maxn];int main(){scanf("%d",&n);for(int i=1;i<=n;i++){scanf("%d",&p[i]);}dp[1]=0;for(int i=1;i<=n;i++){dp[i+1]=(2*dp[i]-dp[p[i]]+2+mod)%mod;}printf("%lld\n",(dp[n+1]+mod)%mod);return 0;
}

转载于:https://www.cnblogs.com/whdsunny/p/10514150.html

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