#424

A. Unimodal Array

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).

不超过一个峰→yes

#include
#include
using namespace std;
#define maxn 105
int f[maxn];
int com[maxn];
int main()
{
    int n, i = 0, j = 0, top = 0, k = 0, flag = 1, mid = 0;
    scanf("%d", &n);
    scanf("%d", &f[0]);
    top = f[0];
    mid = 0;
    for(i = 1; i < n; ++i){
        scanf("%d", &f[i]);
        if(top <= f[i]){
            top = f[i];
            mid = i;
        }
    }
    for(i = 1; i <= mid; ++i){
        if(f[i] <= f[i-1] && f[i] != top) flag = 0;
    }
    for(i = mid+1; i < n; ++i){
        if(f[i] >= f[i-1])flag = 0;
    }
    if(flag)printf("YES\n");
    else    printf("NO\n");
}

B. Keyboard Layouts

There are two popular keyboard layouts in Berland, they differ only in letters positions. All the other keys are the same. In Berland they use alphabet with 26 letters which coincides with English alphabet.

You are given two strings consisting of 26 distinct letters each: all keys of the first and the second layouts in the same order.

You are also given some text consisting of small and capital English letters and digits. It is known that it was typed in the first layout, but the writer intended to type it in the second layout. Print the text if the same keys were pressed in the second layout.

Since all keys but letters are the same in both layouts, the capitalization of the letters should remain the same, as well as all other characters.

Input
The first line contains a string of length 26 consisting of distinct lowercase English letters. This is the first layout.

The second line contains a string of length 26 consisting of distinct lowercase English letters. This is the second layout.

The third line contains a non-empty string s consisting of lowercase and uppercase English letters and digits. This is the text typed in the first layout. The length of s does not exceed 1000.

Output
Print the text if the same keys were pressed in the second layout.

Examples
input
qwertyuiopasdfghjklzxcvbnm
veamhjsgqocnrbfxdtwkylupzi
TwccpQZAvb2017
output
HelloVKCup2017

input
mnbvcxzlkjhgfdsapoiuytrewq
asdfghjklqwertyuiopzxcvbnm
7abaCABAABAcaba7
output
7uduGUDUUDUgudu7

耗时的做法↓ 可以用hash，或map

#include
#include
#include
using namespace std;
#define maxn 1005
int main()
{
    char a[maxn], b[maxn], str[maxn];
    int i, j;
    scanf("%s",a);
    scanf("%s",b); 
    scanf("%s",str);
    for(i = 0; i < strlen(str); ++i){
        if((str[i] >= 'a' && str[i] <= 'z') || (str[i] >= 'A' && str[i] <= 'Z')){
            for(j = 0; j < 26; ++j){
                if(a[j] == str[i]){
                    printf("%c",b[j]);
                } 
                else if((a[j] - 32) == str[i]){
                    printf("%c",b[j]-32);
                }
            }
        }   
        else{
            printf("%c", str[i]);
        }
    }
    printf("\n");
}

//待补充