【Codeforces 349B】Color the Fence

【链接】 我是链接,点我呀:)
【题意】

让你组成一个只由1~9组成的数字 每个数字需要的paint数字给定。 让你组成一个最大的数字,且所有数字的paint的总和不超过v.

【题解】

先求出a中的最小值mi 最后的长度显然就是a/mi啦 然后从高位到低位,优先让高位优先选择大的数字就好. (判断这一位能否为i的条件就是,后面的所有位置全都选择mi 看看会不会超过剩余的paint数量就好

【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 = 10;
    static int L = (int)1e6;
    static class Task{
        
        int v,mi;
        int a[],b[];
        
        public void dfs(int dep,int rest) {
        	if (dep==0) return;
        	for (int i = 9;i >= 1;i--) {
        		if (rest-a[i]-(dep-1)*mi>=0) {
        			b[dep] = i;
        			dfs(dep-1,rest-a[i]);
        			return;
        		}
        	}
        }
        public void solve(InputReader in,PrintWriter out) {
        	a = new int[N+10];
        	b = new int[L+10];
        	v = in.nextInt();
        	for (int i = 1;i <= 9;i++) a[i] = in.nextInt();
        	mi = a[1];
        	for (int i = 1;i <= 9;i++)  mi = Math.min(a[i], mi);
        	int len = v/mi;
        	if (len==0) 
        		out.println(-1);
        	else {
        		dfs(len,v);
        		for (int i = len;i >= 1;i--) out.print(b[i]);
        	}
        }
    }

    

    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 @ 2019-02-18 17:03  AWCXV  阅读(205)  评论(0编辑  收藏  举报