國慶練習1

阿新 • • 發佈：2018-10-06

hide proc 開始 single ons ets word pen lower

Romaji CF 1008A

Vitya has just started learning Berlanese language. It is known that Berlanese uses the Latin alphabet. Vowel letters are "a", "o", "u", "i", and "e". Other letters are consonant.

In Berlanese, there has to be a vowel after every consonant, but there can be any letter after any vowel. The only exception is a consonant "n"; after this letter, there can be any letter (not only a vowel) or there can be no letter at all. For example, the words "harakiri", "yupie", "man", and "nbo" are Berlanese while the words "horse", "king", "my", and "nz" are not.

Help Vitya find out if a word $s$

Input

The first line of the input contains the string $s$

Output

Print "YES" (without quotes) if there is a vowel after every consonant except "n", otherwise print "NO".

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

Examples

Input

sumimasen

Output

YES

Input

ninja

Output

YES

Input

codeforces

Output

NO

Note

In the first and second samples, a vowel goes after each consonant except "n", so the word is Berlanese.

In the third sample, the consonant "c" goes after the consonant "r", and the consonant "s" stands on the end, so the word is not Berlanese.

題目意思：元音字符有a e i o u,輔音字符是除了元音字符以外的字母。

輔音字母後面只能跟元音字母，除了輔音字母n，它後面可以跟任意字符或者不跟字符；

元音字符後面可以跟任意字符。

給你一個字符串，如果滿足以上條件，則輸出YES，否則輸出NO。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <string>
 6 #define ll long long int
 7 char s[110];
 8 char a[6]="aeiou";
 9 int judge(char c)///判斷元音的被調函數
10 {
11     int i;
12     for(i=0;i<6;i++)
13     {
14         if(c==a[i])
15         {
16             return 1;
17             break;
18         }
19     }
20     return 0;
21 }
22 int main()
23 {
24   int i,n,len,flag;
25   gets(s);
26   len= strlen(s);
27   flag=0;
28   for(i=0;i<len;i++)
29   {
30       if(s[i]==‘n‘)
31       {
32           continue;
33       }
34       if(!judge(s[i]))
35       {
36           if(i==len-1)
37           {
38               flag=1;
39           }
40           if(s[i+1]==‘n‘)
41           {
42               flag=1;
43               break;
44           }
45           else if(!judge(s[i+1]))
46           {
47               flag=1;
48               break;
49           }
50       }
51   }
52   if(!flag)
53   {
54       printf("YES\n");
55   }
56   else
57   {
58       printf("NO\n");
59   }
60   return 0;
61 }

Game Shopping CF1009A

Maxim wants to buy some games at the local game shop. There are $n$

Maxim has a wallet which can be represented as an array of integers. His wallet contains $m$

Games in the shop are ordered from left to right, Maxim tries to buy every game in that order.

When Maxim stands at the position $i$

Maxim buys the $i$

For example, for array $c = [2, 4, 5, 2, 4]$

Your task is to get the number of games Maxim will buy.

Input

The first line of the input contains two integers $n$

The second line of the input contains $n$

The third line of the input contains $m$

Output

Print a single integer — the number of games Maxim will buy.

Examples

Input

5 4
2 4 5 2 4
5 3 4 6

Output

Input

5 2
20 40 50 20 40
19 20

Output

Input

6 4
4 8 15 16 23 42
1000 1000 1000 1000

Output

Note

The first example is described in the problem statement.

In the second example Maxim cannot buy any game because the value of the first bill in his wallet is smaller than the cost of any game in the shop.

In the third example the values of the bills in Maxim‘s wallet are large enough to buy any game he encounter until he runs out of bills in his wallet.

題目意思：商店裏有n個遊戲,主人公的錢包裏有m張鈔票，之後依次給出每個遊戲的價格和錢包中的鈔票價值。如果當前鈔票價值能買當前遊戲，就會買遊戲用掉鈔票，之後使用下一張鈔票；不能買就換成下一個遊戲，直到能夠買為止，求最多能買多少遊戲。

解題思路：直接模擬就可以了。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 int c[1010];
 6 int a[1010];
 7 int main()
 8 {
 9     int n,m,i,j;
10     int ans;
11     scanf("%d%d",&n,&m);
12     for(i=1; i<=n; i++)
13     {
14         scanf("%d",&c[i]);
15     }
16     for(i=1; i<=m; i++)
17     {
18         scanf("%d",&a[i]);
19     }
20     i=1;
21     j=1;
22     ans=0;
23     while(i<=n)
24     {
25         if(a[j]>=c[i])
26         {
27             j++;
28             ans++;
29         }
30         if(j==m+1)
31         {
32             break;
33         }
34         i++;
35     }
36     printf("%d\n",ans);
37     return 0;
38 }

Reorder the Array CF 1008C

You are given an array of integers. Vasya can permute (change order) its integers. He wants to do it so that as many as possible integers will become on a place where a smaller integer used to stand. Help Vasya find the maximal number of such integers.

For instance, if we are given an array $[10, 20, 30, 40]$

Help Vasya to permute integers in such way that the number of positions in a new array, where integers are greater than in the original one, is maximal.

Input

The first line contains a single integer $n$

The second line contains $n$

Output

Print a single integer — the maximal number of the array‘s elements which after a permutation will stand on the position where a smaller element stood in the initial array.

Examples

Input

7
10 1 1 1 5 5 3

Output

Input

5
1 1 1 1 1

Output

Note

In the first sample, one of the best permutations is $[1, 5, 5, 3, 10, 1, 1]$

In the second sample, there is no way to increase any element with a permutation, so the answer is 0.

題目意思：給你一個數組，讓你重新排列，使得當前位置的數比原來的數大的位置最多有多少個。

解題思路：這道題應該怎麽想呢？我在看著一道題的時候，突然想到了一個成語，田忌賽馬！和這道題倒是很像，但這是自己的馬之間互相比賽，問最多贏幾場！但這道題還沒有上升到這種博弈的高度，其實換一下思路和好搞，就是看看那些小的數能否換成比它大的數嘛！我開始的想法是將原來數組從大到小排序，看看讓較大的化成較小的，同時用vis數組控制只能交換一次，得到了下面的代碼：

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 int a[100010];
 6 int vis[100010];
 7 int my_cmp(int x,int y)
 8 {
 9     return x>y;
10 }
11 int main()
12 {
13     int n,i,j,ans,flag;
14     scanf("%d",&n);
15     for(i=0;i<n;i++)
16     {
17         scanf("%d",&a[i]);
18     }
19     sort(a,a+n,my_cmp);
20     ans=0;
21     for(i=0;i<n;i++)
22     {
23         flag=0;
24         for(j=i+1;j<n;j++)
25         {
26             if(vis[j])
27             {
28                 continue;
29             }
30             if(a[i]>a[j]&&!vis[j])
31             {
32                 ans++;
33                 flag=1;
34                 vis[j]=1;
35                 break;
36             }
37         }
38         if(!flag)
39         {
40             break;
41         }
42     }
43     printf("%d\n",ans);
44     return 0;
45 }

View Code

但這個代碼效率並不高，其實可以使用雙指針來做。變成從小到大排序，優先使用最小的數進行比較。用兩個指針來遍歷，指針i與指針j，一開始指針i指在第1個數，指針j指在第2個數，如果a[j]>a[i]，i指針向後移一位，ans++;否則，i指針還是指在原來的位置；j指針向後移，直到j>n。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 int a[100010];
 6 int vis[100010];
 7 int my_cmp(int x,int y)
 8 {
 9     return x>y;
10 }
11 int main()
12 {
13     int n,i,j,ans,flag;
14     scanf("%d",&n);
15     for(i=0;i<n;i++)
16     {
17         scanf("%d",&a[i]);
18     }
19     sort(a,a+n,my_cmp);
20     ans=0;
21     for(i=0;i<n;i++)
22     {
23         flag=0;
24         for(j=i+1;j<n;j++)
25         {
26             if(vis[j])
27             {
28                 continue;
29             }
30             if(a[i]>a[j]&&!vis[j])
31             {
32                 ans++;
33                 flag=1;
34                 vis[j]=1;
35                 break;
36             }
37         }
38         if(!flag)
39         {
40             break;
41         }
42     }
43     printf("%d\n",ans);
44     return 0;
45 }

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hide proc 開始 single ons ets word pen lower Romaji CF 1008A Vitya has just started learning Berlanese language. It is known that Berlanes

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