Codeforces 500 A. New Year Transportation

最新推荐文章于 2021-09-14 23:25:19 发布

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最新推荐文章于 2021-09-14 23:25:19 发布

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本文介绍了一个新年运输系统问题，该系统由一系列连接不同单元格的门户组成。文章探讨了如何从初始位置到达目标位置的可能性，并提供了一段示例代码来解决这一问题。

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New Year is coming in Line World! In this world, there are n cells numbered by integers from 1 to n, as a 1 × n board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.

So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of n - 1positive integers a₁, a₂, ..., a_n - 1. For every integer i where 1 ≤ i ≤ n - 1 the condition 1 ≤ a_i ≤ n - i holds. Next, he made n - 1portals, numbered by integers from 1 to n - 1. The i-th (1 ≤ i ≤ n - 1) portal connects cell i and cell (i + a_i), and one can travel from cell i to cell (i + a_i) using the i-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (i + a_i) to cell i using the i-th portal. It is easy to see that because of condition 1 ≤ a_i ≤ n - i one can't leave the Line World using portals.

Currently, I am standing at cell 1, and I want to go to cell t. However, I don't know whether it is possible to go there. Please determine whether I can go to cell t by only using the construted transportation system.

Input

The first line contains two space-separated integers n (3 ≤ n ≤ 3 × 10⁴) and t (2 ≤ t ≤ n) — the number of cells, and the index of the cell which I want to go to.

The second line contains n - 1 space-separated integers a₁, a₂, ..., a_n - 1 (1 ≤ a_i ≤ n - i). It is guaranteed, that using the given transportation system, one cannot leave the Line World.

Output

If I can go to cell t using the transportation system, print "YES". Otherwise, print "NO".

Examples
inputCopy
8 4
1 2 1 2 1 2 1
outputCopy
YES
inputCopy
8 5
1 2 1 2 1 1 1
outputCopy
NO

Note

In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4.

In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit.

思路：

在第一个示例中，已访问的单元格是：1, 2, 4；因此，我们能够成功访问单元格 4。

在第二个示例中，可能访问的单元格是：1, 2, 4, 6, 7, 8；因此，我们无法访问想要前往的单元格 5。

代码：

#include<stdio.h>
#include<iostream>
#include<algorithm>
int a[30001];
int b[30001];
using namespace std;
int main(){
	int n,t,j=1;
	cin>>n>>t;
	int i;
	for(i=1;i<=n-1;i++)
	   cin>>a[i];
	for(i=1;;)
	{
		b[j++]=i+a[i];
		i=i+a[i];
		if(i>t)
		  break;
	}
	int flag=0;
	for(i=1;i<j;i++)
	{
		if(b[i]==t)
		 {
		 	flag=1;
		 	break;
		 }
	}
	if(flag==1)
	  cout<<"YES"<<endl;
	else
	  cout<<"NO"<<endl;
	return 0;	
}