• 走房间


    题目网址:http://codeforces.com/problemset/problem/407/B

    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 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 Input

    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:开辟一个数组q[],q[i]表示从第i个房间的第二个门走到p[i]房间然后回到i房间的路程
    #include <iostream>
    #include <algorithm>
    #include <cstring>
    #include <cstdio>
    #include <cmath>
    
    using namespace std;
    int p[1005];
    int q[1005];
    
    int main()
    {
        int n,i,j;
        long long sum,sum1;
        while(cin>>n)
        {
            sum=2;
            memset(q,0,sizeof(q));
            for(i=1;i<=n;i++)
            cin>>p[i];
            q[1]=2;
            for(i=2;i<=n;i++)
            {
                sum1=0;
                if(p[i]<i)
                {
                   for(j=p[i];j<i;j++)
                   {
                       sum1+=q[j];
                   }
                   sum1=sum1+2;
                   if(sum1>=1000000007)
                   sum1=sum1%1000000007;
                   q[i]=sum1;
                   sum+=sum1;
                }
                else
                {
                    q[i]=2;
                    sum=sum+2;
                }
                if(sum>=1000000007)
                sum=sum%1000000007;
            }
            cout<<sum<<endl;
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/chen9510/p/4982360.html
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