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【Codeforces 979B】Treasure Hunt

范霄
2023-12-01

【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());
        }
    }
}

转载于:https://www.cnblogs.com/AWCXV/p/10485547.html

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