【HDU】 4722 Good Numbers

Good Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3679    Accepted Submission(s): 1166

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.

   
   
   
    
  
  

 

题解：这一题可以用规律来找...首先判断1~a中满足要求的一定是a/10 或a/10-1，如果（a/10）*10~a中有满足题目要求的数，就可以不减1，否则减一。

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;

int T,t;
long long ansa,ansb,ans,a,b;

int solve(long long x)
{
    int t=x%10,f=0,k,i=0;
    x=x-t;
    x=x/10;
    while (x!=0)
    {
        k=x%10;
        x=x-k;
        x=x/10;
        f+=k;
    }
    while (f%10!=0)
    {
        i++;
        f+=1;
    }
    if (i==t) return 2;
    if (t>i) return 1;
    else return 0;
}

int main()
{
    scanf("%d",&T);
    t=T;
    while (T--)
    {
        scanf("%I64d%I64d",&a,&b);
        ansa=a/10;
        if (!solve(a)) ansa--;
        ansb=b/10;
        if (!solve(b)) ansb--;
        ans=ansb-ansa;
        if (solve(a)==2) ans++;
        printf("Case #%d: %I64d\n",t-T,ans);
    }
    return 0;
}