【链接】 我是链接,点我呀:)
【题意】
每次你可以将一个字符变成一个不同于本身的字符.
每个人需要改变n次(且不能不改变)
设每个人的字符串中出现次数最多的字符出现的次数为cnt[0~2]
问你谁的cnt值最大
如果最大的两个cnt相同 输出draw
【题解】
模拟就好
注意这种情况(n=1)
1
aaa
aab
第一个人因为必须要改变一个,所以cnt最大为2
第二个人则为3
【代码】
import java.io.*;
import java.util.*;
public class Main {
static InputReader in;
static PrintWriter out;
public static void main(String[] args) throws IOException{
//InputStream ins = new FileInputStream("E:\\rush.txt");
InputStream ins = System.in;
in = new InputReader(ins);
out = new PrintWriter(System.out);
//code start from here
new Task().solve(in, out);
out.close();
}
static int N = 50000;
static class Task{
int n;
String s[] = new String[3];
int ma[] = new int[3];
int cnt[][] = new int[3][500];
int len;
TreeMap dic = new TreeMap();
public void solve(InputReader in,PrintWriter out) {
n = in.nextInt();
for (int i = 0;i < 3;i++) s[i] = in.next();
for (int i = 0;i < 3;i++) {
for (int j = 0;j < (int)s[i].length();j++) {
cnt[i][(int)s[i].charAt(j)]++;
}
for (int j = 'a';j <= 'z';j++)
ma[i] = Math.max(ma[i], cnt[i][j]);
for (int j = 'A';j <='Z';j++)
ma[i] = Math.max(ma[i], cnt[i][j]);
if (ma[i]==(int)s[i].length()) {
if (n%2==1) {
ma[i]--;
}
}else {
ma[i] = Math.min(ma[i]+n, s[i].length());
}
}
String s[] = {"Kuro","Shiro","Katie"};
if(ma[0]>ma[1]) {
int x = ma[0];ma[0] = ma[1];ma[1] = x;
String t = s[0];s[0] = s[1];s[1] = t;
}
if(ma[1]>ma[2]) {
int x = ma[1];ma[1] = ma[2];ma[2] = x;
String t = s[1];s[1] = s[2];s[2] = t;
}
if(ma[0]>ma[1]) {
int x = ma[0];ma[0] = ma[1];ma[1] = x;
String t = s[0];s[0] = s[1];s[1] = t;
}
if (ma[1]==ma[2]) {
out.println("Draw");
}else {
out.println(s[2]);
}
}
}
static class InputReader{
public BufferedReader br;
public StringTokenizer tokenizer;
public InputReader(InputStream ins) {
br = new BufferedReader(new InputStreamReader(ins));
tokenizer = null;
}
public String next(){
while (tokenizer==null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(br.readLine());
}catch(IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
}