Long Path//CodeForces - 407B//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
2
1 2
Output
4
Input
4
1 1 2 3
Output
20
Input
5
1 1 1 1 1
Output
62
题意
奇数次按照给出的pi传送，偶数次保证到下一个房间，问走出n个房间穿过门的次数
链接:https://vjudge.net/contest/349029#problem/G

思路

分析见代码处，找出能确保穿出当前房间的次数再慢慢推即可

代码

#include <cstdio>
#include<algorithm>
using namespace std;
const int inf = 0x3f3f3f3f;
const int mod = 1e9+7;
int dp[100010];
int num[100010];
int p[100010];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&p[i]);
    dp[1]=2;                    //两次才能从第一个房间的正常出口出去
    num[1]=2;                   //需要至少两次才能确保到达2房间
    for(int i=2;i<=n;i++){
        dp[i]=num[i-1]-num[p[i]-1]+2;  //第二次到达该房间的次数=确保穿过上一个房间所需的次数
        if(dp[i]<0) dp[i]+=mod;        //-奇数次进入该房间后去到下一个房间离开所需次数+2次过门的次数
        num[i]=dp[i]+num[i-1];         //确保穿过此房间的次数
        num[i]%=mod;
    }
    printf("%d\n",num[n]);
    return 0;
}

注意

dp[i]可能为负数，别忘了加mod

来源：CSDN

作者：salty_fishman

链接：https://blog.csdn.net/salty_fishman/article/details/104216279

标签

portal