CUC-SUMMER-8-A

A - New Year Transportation

CodeForces - 500A

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 - 1 positive integers a1, a2, ..., an - 1. For every integer i where 1 ≤ i ≤ n - 1 the condition 1 ≤ ai ≤ n - i holds. Next, he made n - 1 portals, numbered by integers from 1 to n - 1. The i-th (1 ≤ i ≤ n - 1) portal connects cell i and cell (i + ai), and one can travel from cell i to cell (i + ai) using the i-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (i + ai) to cell i using the i-th portal. It is easy to see that because of condition 1 ≤ ai ≤ 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, ..., an - 1 (1 ≤ ai ≤ 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".

Example
Input
8 4
1 2 1 2 1 2 1
Output
YES
Input
8 5
1 2 1 2 1 1 1
Output
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.

题意：一个一维的轴上，你在1位置处，每一个坐标有一个权值代表向前移动的距离，你有一个目的地，问能不能到达

解法：水题，从1开始模拟即可，如果到达目的坐标或者超过目的坐标则循环终止。

代码：

#include<iostream>
using namespace std;
int a[30005];
int main()
{
    int n,t,s;
    cin>>n>>t;
    for(int i=1;i<n;i++)
        cin>>a[i];
    s=1;
    while(1){
        if(s==t){
            cout<<"YES"<<endl;
            break;
        }
        if(s>t){
            cout<<"NO"<<endl;
            break;
        }
        s+=a[s];
    }
    return 0;
}