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 <= 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

 

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。

 

 1 #include<cstdio>

 2 #include<cstring>

 3 #include<iostream>

 4 #include<algorithm>

 5 using namespace std;

 6 int keng(long long x,int left)

 7 {

 8     int n=0,i,ans;

 9     while(x)

10     {

11         n+=x%10;

12         n%=10;

13         x/=10;

14     }

15     ans=left/10;

16     i=10-(n+ans)%10;

17         if(i==10)

18             i=0;

19     if(left%10>=i)

20         ans++;

21     return ans;

22 }

23 bool check(long long a)

24 {

25     int n=0;

26     while(a)

27     {

28         n+=a%10;

29         n%=10;

30         a/=10;

31     }

32     if(n%10==0)

33         return true;

34     else return false;

35 }

36 int main()

37 {

38     int t,i,Case,shu;

39     long long a,b;

40     long long Ca,Cb;

41     long long left;

42     scanf("%d",&t);

43     // a=1;

44     for(Case=1;Case<=t;Case++)

45     {

46         // b=Case;

47         scanf("%I64d%I64d",&a,&b);

48         Ca=a/100*10;left=a%100;

49         Ca=keng(Ca/10,left)+Ca;

50         if (check(a)) Ca--;

51         Cb=b/100*10;left=b%100;

52         Cb=keng(Cb/10,left)+Cb;

53         printf("Case #%d: %I64d\n",Case,Cb-Ca);

54     }

55     return 0;

56 } 

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