3 2

1 3 2

3 1

1 2 3

5 2

1 3 2 4 5

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;const int maxn=100000+1000;int ans[maxn];int main(){    int n,k;    ans[1]=1;    scanf("%d%d",&amp;n,&amp;k);    if(k==1)    {        for(int i=1;i   <br><br><p></p>   </cstring></algorithm></cstdio></iostream>

We all know that a superhero can transform to certain other superheroes. But not all Superheroes can transform to any other superhero. A superhero with name ss can transform to another superhero with name tt if ss can be made equal to tt by changing any vowel in ss to any other vowel and any consonant in ss to any other consonant. Multiple changes can be made.

In this problem, we consider the letters ''a'', ''e'', ''i'', ''o'' and ''u'' to be vowels and all the other letters to be consonants.

Given the names of two superheroes, determine if the superhero with name ss can be transformed to the Superhero with name tt.

The first line contains the string ss having length between 11 and 10001000, inclusive.

The second line contains the string tt having length between 11 and 10001000, inclusive.

Both strings ss and tt are guaranteed to be different and consist of lowercase English letters only.

Output "Yes" (without quotes) if the superhero with name ss can be transformed to the superhero with name tt and "No" (without quotes) otherwise.

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

a
u

Yes

abc
ukm

Yes

akm
ua

No

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 
 5 int vow(char a,char b)
 6 {
 7     if(a == b)
 8     return 1;
 9     else if(a == ''a'' || a == ''o'' || a == ''e'' || a == ''i'' || a == ''u'')
10     {
11         if(b == ''a'' || b == ''o'' || b == ''e'' || b == ''i'' || b == ''u'')//元元
12         return 1;
13         else//元辅
14         return 0;
15     }
16     else if(b == ''a'' || b == ''o'' || b == ''e'' || b == ''i'' || b == ''u'')//辅元
17     return 0;
18     //辅辅
19     return 1;
20 }
21 int main()
22 {
23     string s,t;
24     while(cin >> s >> t)
25     {
26         int sl = s.size();
27         int tl = t.size();
28         if(sl != tl)
29         {
30             cout << "No" << endl;
31             continue;
32         }
33         bool flag = 0;
34         for(int i = 0;i < sl;i++)
35         {
36             if(vow(s[i],t[i]) == 0)
37             {
38                 cout << "No" << endl;
39                 flag = 1;
40                 break;
41             }
42         }
43         if(flag == 0)
44         cout << "Yes" << endl;
45     }
46     return 0;
47 }

Vasya is reading a e-book. The file of the book consists of nn pages, numbered from 11 to nn. The screen is currently displaying the contents of page xx, and Vasya wants to read the page yy. There are two buttons on the book which allow Vasya to scroll dd pages forwards or backwards (but he cannot scroll outside the book). For example, if the book consists of 1010 pages, and d=3d=3, then from the first page Vasya can scroll to the first or to the fourth page by pressing one of the buttons; from the second page — to the first or to the fifth; from the sixth page — to the third or to the ninth; from the eighth — to the fifth or to the tenth.

Help Vasya to calculate the minimum number of times he needs to press a button to move to page yy.

The first line contains one integer tt (1≤t≤1031≤t≤103) — the number of testcases.

Each testcase is denoted by a line containing four integers nn, xx, yy, dd (1≤n,d≤1091≤n,d≤109, 1≤x,y≤n1≤x,y≤n) — the number of pages, the starting page, the desired page, and the number of pages scrolled by pressing one button, respectively.

Print one line for each test.

If Vasya can move from page xx to page yy, print the minimum number of times he needs to press a button to do it. Otherwise print −1−1.

In the first test case the optimal sequence is: 4→2→1→3→54→2→1→3→5.

In the second test case it is possible to get to pages 11 and 55.

In the third test case the optimal sequence is: 4→7→10→13→16→194→7→10→13→16→19.

翻到头只能从头翻，翻到尾只能从尾倒着翻。

代码:

 1 //A
 2 #include<bits/stdc++.h>
 3 using namespace std;
 4 typedef long long ll;
 5 const int maxn=1e5+10;
 6 const int inf=0x3f3f3f3f;
 7 
 8 int main()
 9 {
10     int t;
11     cin>>t;
12     while(t--){
13         int n,x,y,d,ans=inf;
14         cin>>n>>x>>y>>d;
15         if(abs(y-x)%d==0) {ans=min(ans,abs(y-x)/d);}
16         if((y-1)%d==0) {ans=min(ans,x/d+(y-1)/d+(x%d==0?0:1));}
17         if((n-y)%d==0) {ans=min(ans,(n-x)/d+(n-y)/d+((n-x)%d==0?0:1));}
18         if(ans==inf) cout<<-1<<endl;
19         else cout<<ans<<endl;
20     }
21 }