【Codeforces 348A】Mafia

【链接】 我是链接,点我呀:)
【题意】

每轮游戏都要有一个人当裁判,其余n-1个人当玩家 给出每个人想当玩家的次数ai 请你求出所需要最少的玩游戏的轮数 使得每个人都能满足他们当玩家的要求.

【题解】

```cpp /* * 如果有x轮的话 * 理论上每个人都能玩x次 * 但是必须有x人当supervisor * x次的话,每个人能当supervisor的次数为x-a[i] * n*x-∑a[i]就是所有人除去必须的能打supervisor的次数 * 如果n*x-∑a[i] >= x 那么就ok * (n-1)*x-∑a[i]>=0就好 *x>=∑a[i] / (n-1) *要注意x需要大于等于max{a[i]} */ ```

【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 = 100000;
    static class Task{
        /*
         * 如果有x轮的话
         * 理论上每个人都能玩x次
         * 但是必须有x人当supervisor
         * x次的话,每个人能当supervisor的次数为x-a[i]
         * n*x-∑a[i]就是所有人除去必须的能打supervisor的次数
         * 如果n*x-∑a[i] >= x 那么就ok
         * (n-1)*x-∑a[i]>=0就好
         *x>=∑a[i] / (n-1)
         */
    	int n;
    	int a[];
        
        public void solve(InputReader in,PrintWriter out) {
        	n = in.nextInt();
        	a = new int[N+10];
        	int ma = 0;
        	for (int i = 1;i <= n;i++) {
        		a[i] = in.nextInt();
        		ma = Math.max(ma,a[i]);
        	}
        	long ans;
        	long sum = 0;
        	for (int i = 1;i <= n;i++) sum += a[i];
	        	if (sum%(n-1)==0) {
	        		ans = sum/(n-1);
	        	}else {
	        		ans = sum/(n-1) + 1;
	        	}
	        ans = Math.max(ans, ma);
	        out.println(ans);
        }
    }

    

    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());
        }
    }
}
posted @ 2019-02-23 21:18  AWCXV  阅读(692)  评论(0编辑  收藏  举报