HDU 4722 Good Numbers

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4722

Good Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 422    Accepted Submission(s): 146

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 <= 10¹⁸).

 

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

 

题目大意：找出给定范围内数位和能被10整除的数的数目

 

解题思路：不难发现，以100个数为一组，每组有10个数满足条件，对于给定的数，先找有多少组，对剩余的数暴搜一遍就能得到结果。

 

 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 using namespace std;
 int keng(long long x,int left)
 {
     int n=,i,ans;
     while(x)
     {
         n+=x%;
         n%=;
         x/=;
     }
     ans=left/;
     i=-(n+ans)%;
         if(i==)
             i=;
     if(left%>=i)
         ans++;
     return ans;
 }
 bool check(long long a)
 {
     int n=;
     while(a)
     {
         n+=a%;
         n%=;
         a/=;
     }
     if(n%==)
         return true;
     else return false;
 }
 int main()
 {
     int t,i,Case,shu;
     long long a,b;
     long long Ca,Cb;
     long long left;
     scanf("%d",&t);
     // a=1;
     for(Case=;Case<=t;Case++)
     {
         // b=Case;
         scanf("%I64d%I64d",&a,&b);
         Ca=a/*;left=a%;
         Ca=keng(Ca/,left)+Ca;
         if (check(a)) Ca--;
         Cb=b/*;left=b%;
         Cb=keng(Cb/,left)+Cb;
         printf("Case #%d: %I64d\n",Case,Cb-Ca);
     }
     return ;
 }