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

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 ;

 }

巴特西

HDU 4722 Good Numbers

Good Numbers

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