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  • Codeforces Round #239 (Div. 1)

    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 numberpi, 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 test(s)
    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:Vasya在一个由各种房间组成的迷宫里,房间的编号为(1~n),开始Vasya在1号房间,每个房间都有两个门,第一扇门通往下一个房间,第二扇门通往任意一个房间,Vasya决定每到一个房间就在该房间标记一次,如果标记数为奇数就打开第二扇门,否则打开第一扇门,问Vasya走出迷宫需要的步数。
    思路:dp[i]表示偶数次到第i个房间的步数,则状态转移方程为dp[i] = dp[i-1]+1+dp[i-1]-dp[a[i]-1]+1;
     1 #include <stdio.h>
     2 #include <iostream>
     3 #include <string.h>
     4 using namespace std;
     5 const int N=1002;
     6 const int MOD=1000000007;
     7 long long dp[N],a[N];
     8 int main()
     9 {
    10     int n;
    11     while(~scanf("%d",&n))
    12     {
    13         for (int i = 1; i <= n; i++)
    14             cin>>a[i];
    15         dp[0] = -1;
    16         dp[1] = 1;
    17         for (int i = 2; i <= n; i++)
    18         {
    19             dp[i] = (dp[i-1]+1+dp[i-1]-dp[a[i]-1]+1)%MOD;
    20         }
    21         cout<<((dp[n]+1)%MOD+MOD)%MOD<<endl;
    22     }
    23     return 0;
    24 }
    View Code
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  • 原文地址:https://www.cnblogs.com/lahblogs/p/3635743.html
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