Codeforces Round #482 (Div. 2) 979B - Treasure Hunt [思维]

B. Treasure Hunt

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

After the big birthday party, Katie still wanted Shiro to have some more fun. Later, she came up with a game called treasure hunt. Of course, she invited her best friends Kuro and Shiro to play with her.

The three friends are very smart so they passed all the challenges very quickly and finally reached the destination. But the treasure can only belong to one cat so they started to think of something which can determine who is worthy of the treasure. Instantly, Kuro came up with some ribbons.

A random colorful ribbon is given to each of the cats. Each color of the ribbon can be represented as an uppercase or lowercase Latin letter. Let's call a consecutive subsequence of colors that appears in the ribbon a subribbon. The beauty of a ribbon is defined as the maximum number of times one of its subribbon appears in the ribbon. The more the subribbon appears, the more beautiful is the ribbon. For example, the ribbon aaaaaaa has the beauty of $7$ because its subribbon a appears $7$ times, and the ribbon abcdabc has the beauty of $2$ because its subribbon abc appears twice.

The rules are simple. The game will have $n$ turns. Every turn, each of the cats must change strictly one color (at one position) in his/her ribbon to an arbitrary color which is different from the unchanged one. For example, a ribbon aaab can be changed into acab in one turn. The one having the most beautiful ribbon after $n$ turns wins the treasure.

Could you find out who is going to be the winner if they all play optimally?

Input

The first line contains an integer $n$ ($0\le n\le {10}^9$) — the number of turns.

Next 3 lines contain 3 ribbons of Kuro, Shiro and Katie one per line, respectively. Each ribbon is a string which contains no more than ${10}^5$uppercase and lowercase Latin letters and is not empty. It is guaranteed that the length of all ribbons are equal for the purpose of fairness. Note that uppercase and lowercase letters are considered different colors.

Output

Print the name of the winner ("Kuro", "Shiro" or "Katie"). If there are at least two cats that share the maximum beauty, print "Draw".

Examples

input

Copy

3
Kuroo
Shiro
Katie

output

Copy

Kuro

input

Copy

7
treasurehunt
threefriends
hiCodeforces

output

Copy

Shiro

input

Copy

1
abcabc
cbabac
ababca

output

Copy

Katie

input

Copy

15
foPaErcvJ
mZaxowpbt
mkuOlaHRE

output

Copy

Draw

Note

In the first example, after $3$ turns, Kuro can change his ribbon into ooooo, which has the beauty of $5$, while reaching such beauty for Shiro and Katie is impossible (both Shiro and Katie can reach the beauty of at most $4$, for example by changing Shiro's ribbon into SSiSS and changing Katie's ribbon into Kaaaa). Therefore, the winner is Kuro.

In the fourth example, since the length of each of the string is $9$ and the number of turn is $15$, everyone can change their ribbons in some way to reach the maximal beauty of $9$

 by changing their strings into zzzzzzzzz after 9 turns, and repeatedly change their strings into azzzzzzzz and then into zzzzzzzzz thrice. Therefore, the game ends in a draw.

题意：就是三个字符串，分别进行n次操作（将一个字符变为一个与其不同的字符），问最后谁的相同字符数量最大。。。

思路：首先我说一下，比赛时读错题是真的很凉，就是直接找出现次数最多的字符串，然后判断一下就好了，为什么我非要想成连续的呢，真的好凉，求出字符串出现次数最多的，

当nmax == len&&n == 1时，长度是len - 1；

当n <= len - nmax时， 长度是nmax + n;

否则 总有办法达到长度为len;

代码：

#include<bits/stdc++.h>
#include<map>
#define ll long long

using namespace std;
const int maxn = 1e5+7;
int n, p[3];
char a[maxn];
map<char, int> mp;
map<char, int> :: iterator it;

int find(char s[])
{
    mp.clear();
    int len = strlen(s), nmax = 0;
    for(int i = 0; i < len; i++) mp[s[i]]++;
    for(it = mp.begin(); it != mp.end(); it++)
    nmax = max(nmax, (it->second));
    if(n == 1&&nmax == len) nmax = len - 1;
    else if(len - nmax >= n) nmax += n;
    else nmax = len;
    return nmax;
}



int main()
{
    //freopen("in.txt", "r", stdin);
    while(~scanf("%d",&n))
    {
        for(int i = 0; i < 3; i++) {
                scanf("%s", a);
                p[i] = find(a);
        }
        int nmax = max(p[0], max(p[1], p[2]));
        if((p[0] == nmax&&p[1] == nmax)||(p[0] == nmax&&p[2] == nmax)||(p[1] == nmax&&p[2] == nmax)) puts("Draw");
        else if(p[0] == nmax) puts("Kuro");
        else if(p[1] == nmax) puts("Shiro");
        else if(p[2] == nmax) puts("Katie");
        else puts("Draw");
    }
    return 0;
}