zoukankan      html  css  js  c++  java
  • CodeForces

    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 tby 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".

    Examples

    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.

    题解:模拟更新可以到达的位置,然后加上该位置下标的值

    代码:

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<iostream>
    
    using namespace std;
    
    int main() {
    	int n,m;
    	cin>>n>>m;
    	int a[30005];
    	for(int t=1; t<=n-1; t++) {
    		scanf("%d",&a[t]);
    	}
    	int flag=0;
    	for(int t=1; t<n;) {
    		t=t+a[t];
    		//cout<<t<<endl;
    		if(t==m) {
    			flag=1;
    		}
    	}
    
    	if(flag==1) {
    		cout<<"YES"<<endl;
    	} else {
    		cout<<"NO"<<endl;
    	}
    	return 0;
    }
  • 相关阅读:
    JavaAPI基础(1)
    类与对象、封装、构造方法
    Java语言基础
    Request请求的应用
    Response的应用
    java生成动态验证码
    Servlet的配置
    常见的状态码。。
    简单学习【1】——打包JS
    NodeJS2-2环境&调试----引用系统内置模块,引用第三方模块
  • 原文地址:https://www.cnblogs.com/Staceyacm/p/10781884.html
Copyright © 2011-2022 走看看