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  • 【Codeforces 1036C】Classy Numbers

    【链接】 我是链接,点我呀:)
    【题意】

    让你求出只由3个非0数字组成的数字在[li,ri]这个区间里面有多少个.

    【题解】

    只由3个非0数字组成的数字在1~10^18中只有60W个 dfs处理出来之后排序做个二分查找一下区间里有多少个就好。

    【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 = 700000;
        static class Task{
            TreeSet<Long> myset = new TreeSet<Long>();
            long a[] = new long[N+10];
            int T,n;
            long l,r;
            
            void dfs(long cur,int num,int dep) {
            	if (dep==18) {
            		if (cur>0) myset.add(cur);	
            		return;
            	}
            	if (num<3) {
            		for (int j = 1;j <= 9;j++)
            			dfs(cur*10+j,num+1,dep+1);
            	}
            	dfs(cur*10,num,dep+1);
            }
            
            int get_least_gore(long x) {
            	int l = 1,r = n,temp = 1;
            	while (l<=r) {
            		int mid = (l+r)/2;
            		if (a[mid]>=x) {
            			temp = mid;
            			r = mid - 1;
            		}else l = mid + 1;
            	}
            	return temp;
            }
            
            int get_last_lore(long x) {
            	int l = 1,r = n,temp = 1;
            	while (l<=r) {
            		int mid = (l+r)/2;
            		if (a[mid]<=x) {
            			temp = mid;
            			l = mid + 1;
            		}else r = mid - 1;
            	}
            	return temp;
            }
            
            public void solve(InputReader in,PrintWriter out) {
            	myset.add((long)1e18);
            	dfs(0,0,0);
            	n = myset.size();
            	Iterator<Long> it = myset.iterator();
            	int j = 1;
            	while (it.hasNext()) {
            		a[j] = it.next();
            		j++;
            	}
            	T = in.nextInt();
            	for (int ii = 1;ii <= T;ii++) {
            		l = in.nextLong();r = in.nextLong();
            		int idx1 = get_least_gore(l);
            		int idx2 = get_last_lore(r);
            		out.println(idx2-idx1+1);
            	}
            }
        }
    
        
    
        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());
            }
            
            public long nextLong() {
            	return Long.parseLong(next());
            }
        }
    }
    
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  • 原文地址:https://www.cnblogs.com/AWCXV/p/10518123.html
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