• CF-831A


    A. Unimodal Array
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Array of integers is unimodal, if:

    • it is strictly increasing in the beginning;
    • after that it is constant;
    • after that it is strictly decreasing.

    The first block (increasing) and the last block (decreasing) may be absent. It is allowed that both of this blocks are absent.

    For example, the following three arrays are unimodal: [5, 7, 11, 11, 2, 1], [4, 4, 2], [7], but the following three are not unimodal: [5, 5, 6, 6, 1], [1, 2, 1, 2], [4, 5, 5, 6].

    Write a program that checks if an array is unimodal.

    Input

    The first line contains integer n (1 ≤ n ≤ 100) — the number of elements in the array.

    The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1 000) — the elements of the array.

    Output

    Print "YES" if the given array is unimodal. Otherwise, print "NO".

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

    Examples
    input
    6
    1 5 5 5 4 2
    output
    YES
    input
    5
    10 20 30 20 10
    output
    YES
    input
    4
    1 2 1 2
    output
    NO
    input
    7
    3 3 3 3 3 3 3
    output
    YES
    Note

    In the first example the array is unimodal, because it is strictly increasing in the beginning (from position 1 to position 2, inclusively), that it is constant (from position 2 to position 4, inclusively) and then it is strictly decreasing (from position 4 to position 6, inclusively).

    题意:

    递增,相等,递减只能按顺序出现,否则为“NO”。

    AC代码:

     1 #include<bits/stdc++.h>
     2 using namespace std;
     3 
     4 int cmp(int x,int y){
     5     if(x<y)
     6     return 1;
     7     if(x==y)
     8     return 2;
     9     if(x>y)
    10     return 3;
    11 }
    12 
    13 int main(){
    14     int n,x[1010];
    15     cin>>n;
    16     int t=1;
    17     for(int i=0;i<n;i++){
    18         cin>>x[i];
    19     }
    20     for(int i=1;i<n;i++){
    21         if(cmp(x[i-1],x[i])<t){
    22             cout<<"NO"<<endl;
    23             return 0;
    24         }
    25         else{
    26             t=cmp(x[i-1],x[i]);
    27         }
    28     }
    29     cout<<"YES"<<endl;
    30     return 0;
    31 } 
  • 相关阅读:
    HDU 1312 Red and Black(经典DFS)
    POJ 1274 The Perfect Stall(二分图 && 匈牙利 && 最小点覆盖)
    POJ 3041 Asteroids(二分图 && 匈牙利算法 && 最小点覆盖)
    HDU 1016 素数环(dfs + 回溯)
    HDU 1035 Robot Motion(dfs + 模拟)
    vjudge Trailing Zeroes (III) (二分答案 && 数论)
    openjudge 和为给定数(二分答案)
    图的存储
    二分查找
    快速选择算法
  • 原文地址:https://www.cnblogs.com/Kiven5197/p/7229813.html
Copyright © 2020-2023  润新知