codefores 834B The Festive Evening

B. The Festive Evening

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output



It's the end of July – the time when a festive evening is held at Jelly Castle! Guests from all over the kingdom gather here to discuss new trends in the world of confectionery. Yet some of the things discussed here are not supposed to be disclosed to the general
public: the information can cause discord in the kingdom of Sweetland in case it turns out to reach the wrong hands. So it's a necessity to not let any uninvited guests in.

There are 26 entrances in Jelly Castle, enumerated with uppercase English letters from A to Z.
Because of security measures, each guest is known to be assigned an entrance he should enter the castle through. The door of each entrance is opened right before the first guest's arrival and closed right after the arrival of the last guest that should enter
the castle through this entrance. No two guests can enter the castle simultaneously.

For an entrance to be protected from possible intrusion, a candy guard should be assigned to it. There are k such guards in the castle,
so if there are more than k opened doors, one of them is going to be left unguarded! Notice that a guard can't leave his post until
the door he is assigned to is closed.

Slastyona had a suspicion that there could be uninvited guests at the evening. She knows the order in which the invited guests entered the castle, and wants you to help her check whether there was a moment when more than k doors
were opened.

Input

Two integers are given in the first string: the number of guests n and the number of guards k (1 ≤ n ≤ 106, 1 ≤ k ≤ 26).

In the second string, n uppercase English letters s1s2... sn are
given, where si is
the entrance used by the i-th guest.

Output

Output «YES» if at least one door was unguarded during some time, and «NO»
otherwise.

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

Examples

input

5 1
AABBB

output

NO

input

5 1
ABABB

output

YES

Note

In the first sample case, the door A is opened right before the first guest's arrival and closed when the second guest enters the castle. The door B is opened right before the arrival of the third guest, and closed after the fifth one arrives. One guard can
handle both doors, as the first one is closed before the second one is opened.

In the second sample case, the door B is opened before the second guest's arrival, but the only guard can't leave the door A unattended, as there is still one more guest that should enter the castle through this door.

题意

有n个门，k个守卫，每个门有一个序号，每个嘉宾从特定的门进去

每个门开门到关门之间，必须有一个守卫守候，开门从第一个嘉宾进，最后一个从这个门进的嘉宾进去后关门

问问守卫能不能守住

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
using namespace std;
struct node
{
    int stat;
    int endd;
} dp[100];
int flag[100];
int main()
{
    int n,k,m,sum,aflag;
    char a[1000006];
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        memset(flag,0,sizeof(flag));
        scanf("%s",a);
        aflag=0;
        sum=0;
        for(int i=0; i<n; i++)
        {
            m=a[i]-'A';
            if(flag[m]==0)
            {
                dp[m].stat=i;
                dp[m].endd=i;//为了这组样例
                              //26 1
                             //ABCDEFGHIJKLMNOPQRSTUVWXYZ
                flag[m]=1;
            }
            else
            {
                    dp[m].endd=i;
            }
        }
        for(int i=0; i<n; i++)
        {
            m=a[i]-'A';
            if(dp[m].stat==i)
            {
                sum++;//开门需要一门守卫
                if(sum>k)
                {
                    aflag=1;
                    break;
                }
                if(dp[m].endd==i)//说明只有一个嘉宾从这个门进
                {
                    sum--;
                }
            }
            else if(dp[m].stat<i&&dp[m].stat>i)
            {
                if(sum>k)
                {
                    aflag=1;
                    break;
                }
            }
            else if(dp[m].endd==i)
            {
                if(sum>k)
                {
                    aflag=1;
                    break;
                }
                else
                {
                    sum--;//说明最后一个嘉宾已经进去了，这个门需要关了，不需要人来守护了
                }
            }
        }
        if(aflag==0)
            printf("NO\n");
        else
            printf("YES\n");
    }
}