zoukankan      html  css  js  c++  java
  • Codeforces Round #431 (Div. 2)

    A. Odds and Ends
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Where do odds begin, and where do they end? Where does hope emerge, and will they ever break?

    Given an integer sequence a1, a2, ..., an of length n. Decide whether it is possible to divide it into an odd number of non-empty subsegments, the each of which has an odd length and begins and ends with odd numbers.

    A subsegment is a contiguous slice of the whole sequence. For example, {3, 4, 5} and {1} are subsegments of sequence {1, 2, 3, 4, 5, 6}, while {1, 2, 4} and {7} are not.

    Input

    The first line of input contains a non-negative integer n (1 ≤ n ≤ 100) — the length of the sequence.

    The second line contains n space-separated non-negative integers a1, a2, ..., an (0 ≤ ai ≤ 100) — the elements of the sequence.

    Output

    Output "Yes" if it's possible to fulfill the requirements, and "No" otherwise.

    You can output each letter in any case (upper or lower).

    Examples
    Input
    3
    1 3 5
    Output
    Yes
    Input
    5
    1 0 1 5 1
    Output
    Yes
    Input
    3
    4 3 1
    Output
    No
    Input
    4
    3 9 9 3
    Output
    No
    Note

    In the first example, divide the sequence into 1 subsegment: {1, 3, 5} and the requirements will be met.

    In the second example, divide the sequence into 3 subsegments: {1, 0, 1}, {5}, {1}.

    In the third example, one of the subsegments must start with 4 which is an even number, thus the requirements cannot be met.

    In the fourth example, the sequence can be divided into 2 subsegments: {3, 9, 9}, {3}, but this is not a valid solution because 2 is an even number.

     老早的了,一直忘了贴。
    思路:思维题。

    代码

     1 #include<iostream>
     2 #include<string.h>
     3 using namespace std;
     4 int a[150];
     5 int main(){
     6     int n;
     7     cin>>n;
     8     for(int i=0;i<n;i++){
     9         cin>>a[i];
    10     }
    11     if(n%2==0) cout<<"No"<<endl;
    12     else{
    13         if(a[0]%2==1&&a[n-1]%2==1){
    14             cout<<"Yes"<<endl;
    15         }else{
    16             cout<<"No"<<endl;
    17         }
    18     }
    19     return 0;
    20 }
  • 相关阅读:
    Mysql基本操作
    MySql架构
    并发中关键字的语义
    SpringMVC<一> 基本结构与配置
    Tomcat 顶层结构
    Linux 日常常用指令
    JSON 数据格式
    RMAN 增量备份级别说明
    Oracle常用数据库系统表单以及SQL的整理
    使用puTTY或Xshell连接阿里云TimeOut超时
  • 原文地址:https://www.cnblogs.com/ISGuXing/p/7497251.html
Copyright © 2011-2022 走看看