Good Numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 610 Accepted Submission(s): 225

Problem Description

If we sum up every digit of a number and the result can be exactly divided by 10, we say this number is a good number.
You are required to count the number of good numbers in the range from A to B, inclusive.

Input

The first line has a number T (T <= 10000) , indicating the number of test cases.
Each test case comes with a single line with two numbers A and B (0 <= A <= B <= 10 ¹⁸).

Output

For test case X, output "Case #X: " first, then output the number of good numbers in a single line.

Sample Input

    2 1 10 1 20 
  

Sample Output

    Case #1: 0 Case #2: 1 
    
     Hint 
    
 The answer maybe very large, we recommend you to use long long instead of int.

Source

2013 ACM/ICPC Asia Regional Online —— Warmup2

第一种方法：

10-20 19 可以被10整除

20-30 28 可以被10整除

30-40 37 可以被10整除

律应该很显然了，0~10中有一个，10~20中有一个。。。一次的每十个中必有一个。可以自己在纸上画一下，

import java.io.*;
import java.util.*;
public class Main {
	BufferedReader bu;
	PrintWriter pw;
	long a;
	long b,num1,num2;
	int n;
	public static void main(String[] args) throws IOException{
		new Main().work();
	}
	void work() throws IOException{
		bu=new BufferedReader(new InputStreamReader(System.in));
		pw=new PrintWriter(new OutputStreamWriter(System.out),true);
		n=Integer.parseInt(bu.readLine());
		for(int p=1;p<=n;p++){
			String str[]=bu.readLine().split(" ");
			a=Long.parseLong(str[0]);
			b=Long.parseLong(str[1]);
			
			for(long i=a;;i++){
				if(fun(i)){
					num1=i;//寻找可以被10整除的数
					break;
				}
			}
			for(long i=b;;i--){
				if(fun(i)){
					num2=i;//寻找可以被10整除的数
					break;
				}
			}
			pw.print("Case #"+p+": ");
			pw.println(num2/10-num1/10+1);
		}
	}
	//判断每一位的和能否被10整除
	boolean fun(long a){
		long sum=0;
		while(a!=0){
			sum+=a%10;
			a/=10;
		}
		if(sum%10==0)
			return true;
		return false;
	}
}

第二种方法

import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class Main {
	int t;
	long a,b;
	PrintWriter pw;
	long digdit[]=new long[30];
	long[][] dp=new long[30][30];
	public static void main(String[] args) throws IOException{
		new Main().work();
	}
	void work() throws IOException{
		BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
		pw=new PrintWriter(new OutputStreamWriter(System.out),true);
		t=Integer.parseInt(bu.readLine());
		for(int i=1;i<=t;i++){
			pw.print("Case #"+i+": ");
			String str[]=bu.readLine().split(" ");
			a=Long.parseLong(str[0]);
			b=Long.parseLong(str[1]);
			long num1=dit(b);
			long num2=dit(a-1);
			long ans=num1-num2;
			pw.println(ans);
		}
	}
	long dit(long x){
		int pos=0;
		for(int i=0;i<dp.length;i++)
			Arrays.fill(dp[i], -1);
		
		while(x!=0){
			digdit[pos++]=x%10;//将数字的每一位存储到数组中去
			x=x/10;
		}
		
		return DFS(pos-1,0,true);
	}
	
	long DFS(int pos,int index,boolean boo){
		if(pos==-1){
			if(index==0)
				return 1;
			else
				return 0;	
		}
		if(!boo&&dp[pos][index]!=-1){ 
			return dp[pos][index];
		}
		long s=0;
		int nindex;
		long e=boo?digdit[pos]:9;
		for(int i=0;i<=e;i++){
			nindex=(index+i)%10;//判断是否可以被10整除
			s=s+(DFS(pos-1,nindex,(boo&&(i==e))));
		}
		if(!boo) dp[pos][index]=s;
		return s;
	}
}

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