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=0,i,ans;
    while(x)
    {
        n+=x%10;
        n%=10;
        x/=10;
    }
    ans=left/10;
    i=10-(n+ans)%10;
        if(i==10)
            i=0;
    if(left%10>=i)
        ans++;
    return ans;
}
bool check(long long a)
{
    int n=0;
    while(a)
    {
        n+=a%10;
        n%=10;
        a/=10;
    }
    if(n%10==0)
        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=1;Case<=t;Case++)
    {
        // b=Case;
        scanf("%I64d%I64d",&a,&b);
        Ca=a/100*10;left=a%100;
        Ca=keng(Ca/10,left)+Ca;
        if (check(a)) Ca--;
        Cb=b/100*10;left=b%100;
        Cb=keng(Cb/10,left)+Cb;
        printf("Case #%d: %I64d\n",Case,Cb-Ca);
    }
    return 0;
}