Good Number

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

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#include<iostream>
using namespace std;
int pd(long long n)
{
	int sum=0;
	while(n){
		int z=n%10;
		sum+=z;
		n/=10;
	}
	return (sum%10==0)?1:0;
}
int main()
{
	int n;
	cin>>n;
	for(int i=1;i<=n;i++){
		long long a,b;
		cin>>a>>b;
		a--;//if(a==b)的话就无法判断，所以a要-- 
		long long sum1=a/10;//因为每十个有一个，所以可以直接除以10算出整的 
		for(long long j=a/10*10;j<=a;j++){//遍历找出有没有符合的 
			if(pd(j))
				sum1++;
		}
		long long sum2=b/10;
		for(long long j=b/10*10;j<=b;j++){
			if(pd(j))
				sum2++;
		}
		printf("Case #%d: %lld\n",i,sum2-sum1);
	}
	return 0;
}