HDU 4722 Good Numbers

最新推荐文章于 2020-09-27 19:51:09 发布

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最新推荐文章于 2020-09-27 19:51:09 发布

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分类专栏： 15暑假培训 c/c++ 几何&&数论

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本文介绍了一种算法，用于计算指定范围内好数字的数量。所谓好数字是指该数字的所有位数相加能被10整除的数字。文章提供了一个具体的C语言实现示例，并解释了如何通过调整输入范围来提高效率。

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

打表看出来

每十位中有一个最后要判断一下

#include <stdio.h>

int pd(long long a)
{
    long long k=0;
    while(a)
    {
        k+=a%10;
        a/=10;
    }
    if(k%10!=0)
        return 0;
    else return 1;
}

int main()
{
    int T;
    long long res,a,b,sum,t;
    scanf("%d",&T);
    for(int qq=1;qq<=T;qq++)
    {
        scanf("%lld %lld",&a,&b);
        while(!pd(a))a++;
        while(!pd(b))b--;
        if(a>b)res=0;
        else res=b/10-a/10+1;
        printf("Case #%d: %lld\n",qq,res);
    }
    return 0;
}