2020.10.23-vj个人赛补题

 

B - B

 

Polycarp loves lowercase letters and dislikes uppercase ones. Once he got a string s consisting only of lowercase and uppercase Latin letters.

Let A be a set of positions in the string. Let's call it pretty if following conditions are met:

  • letters on positions from A in the string are all distinct and lowercase;
  • there are no uppercase letters in the string which are situated between positions from A (i.e. there is no such j that s[j] is an uppercase letter, and a1 < j < a2 for some a1 and a2 from A).

Write a program that will determine the maximum number of elements in a pretty set of positions.

Input

The first line contains a single integer n (1 ≤ n ≤ 200) — length of string s.

The second line contains a string s consisting of lowercase and uppercase Latin letters.

Output

Print maximum number of elements in pretty set of positions for string s.

Examples

Input
11
aaaaBaabAbA
Output
2
Input
12
zACaAbbaazzC
Output
3
Input
3
ABC
Output
0

Note

In the first example the desired positions might be 6 and 8or 7 and 8. Positions 6 and 7 contain letters 'a', position 8contains letter 'b'. The pair of positions 1 and 8 is not suitable because there is an uppercase letter 'B' between these position.

In the second example desired positions can be 7, 8 and 11. There are other ways to choose pretty set consisting of three elements.

In the third example the given string s does not contain any lowercase letters, so the answer is 0.

题意：求连续的小写字母中不同字母的最大数量

题解：运用set，set可以去掉重复的字母，存到一个字符串数组里，同时记录下长度，找到最长的输出

#include<iostream>
#include<bits/stdc++.h>
using namespace std;
int main()
{
    char  s[1000];
    int n,ct=0;
    set<char>b;
    cin>>n>>s;
    for(int i=0;i<n;i++)
    {
        if(s[i]>='a'&&s[i]<='z')
        {
            b.insert(s[i]);
        }
        else{
            if(ct<b.size())ct=b.size();
            b.clear();
        }
    }
    if(ct<b.size)ct=b.size();
    b.clear();
    cout<<ct<<endl;
}

E - E

 

Nothing is eternal in the world, Kostya understood it on the 7-th of January when he saw partially dead four-color garland.

Now he has a goal to replace dead light bulbs, however he doesn't know how many light bulbs for each color are required. It is guaranteed that for each of four colors at least one light is working.

It is known that the garland contains light bulbs of four colors: red, blue, yellow and green. The garland is made as follows: if you take any four consecutive light bulbs then there will not be light bulbs with the same color among them. For example, the garland can look like "RYBGRYBGRY", "YBGRYBGRYBG", "BGRYB", but can not look like "BGRYG", "YBGRYBYGR" or "BGYBGY". Letters denote colors: 'R' — red, 'B' — blue, 'Y' — yellow, 'G' — green.

Using the information that for each color at least one light bulb still works count the number of dead light bulbs of each four colors.

Input

The first and the only line contains the string s (4 ≤ |s| ≤ 100), which describes the garland, the i-th symbol of which describes the color of the i-th light bulb in the order from the beginning of garland:

  • 'R' — the light bulb is red,
  • 'B' — the light bulb is blue,
  • 'Y' — the light bulb is yellow,
  • 'G' — the light bulb is green,
  • '!' — the light bulb is dead.

The string s can not contain other symbols except those five which were described.

It is guaranteed that in the given string at least once there is each of four letters 'R', 'B', 'Y' and 'G'.

It is guaranteed that the string s is correct garland with some blown light bulbs, it means that for example the line "GRBY!!!B" can not be in the input data.

Output

In the only line print four integers kr, kb, ky, kg — the number of dead light bulbs of red, blue, yellow and green colors accordingly.

Examples

Input
RYBGRYBGR
Output
0 0 0 0
Input
!RGYB
Output
0 1 0 0
Input
!!!!YGRB
Output
1 1 1 1
Input
!GB!RG!Y!
Output
2 1 1 0

Note

In the first example there are no dead light bulbs.

In the second example it is obvious that one blue bulb is blown, because it could not be light bulbs of other colors on its place according to the statements.

题解：每四个一循环，同时记录下每种缺少的字母的个数，我是特判了字符串长度为4的时候，看是否有不存在的字母，若有则直接设为一，长度大于4时，每四个一循环对比看缺少的是哪个，同时另为“！”的值等于相应的字母，继续重复对比循环

#include<bits/stdc++.h>
using namespace std;
#define speed_up ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
int main()
{
   string s,k;
   int r=0,b=0,y=0,g=0;
   cin>>s;
   int m,i,j;
   m=s.size();
   if(m==4)
   {
      for(i=0;i<4;i++)
      {
          if(s[i]=='R')r++;
          else if(s[i]=='B')b++;
          else if(s[i]=='Y')y++;
          else if(s[i]=='G')g++;
      }
      if(r==0)r=1;else r=0;
       if(b==0)b=1;else b=0;
       if(y==0)y=1;else y=0;
       if(g==0)g=1;else g=0;

      cout<<r<<" "<<b<<" "<<y<<" "<<g<<endl;
   }
   else{
   for(i=0;i<m;i++)
   {
       if(i+4<m)
       {
          if(s[i]=='!')
          {
              for(j=i+4;j<m;j+=4)
              {
                  if(s[j]!='!')
                  {
                      if(s[j]=='R')s[i]='R',r++;
                      else if(s[j]=='B')s[i]='B',b++;
                      else if(s[j]=='Y')s[i]='Y',y++;
                      else if(s[j]=='G')s[i]='G',g++;
                      break;
                  }
              }
          }
       }
        if(i-4>=0)
       {
           if(s[i]=='!')
          {
              for(j=i-4;j>=0;j-=4)
              {
                  if(s[j]!='!')
                  {
                      if(s[j]=='R')s[i]='R',r++;
                      else if(s[j]=='B')s[i]='B',b++;
                      else if(s[j]=='Y')s[i]='Y',y++;
                      else if(s[j]=='G')s[i]='G',g++;
                      break;
                  }
              }
          }
       }
   }
   cout<<r<<" "<<b<<" "<<y<<" "<<g<<endl;
}
}

 

posted @ 2020-10-31 11:43  西瓜0  阅读(147)  评论(0)    收藏  举报