hdu4722 Good Numbers 规律题

Good Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3839    Accepted Submission(s): 1224

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
Hint
The answer maybe very large, we recommend you to use long long instead of int.
 

 

开始可以暴力打印一些出来看看，注意0能够被10整除

0 - 100  10

0 - 1000  100

0 - 99 10

0 - 91 10

0 - 90 9

可以发现有近似num / 10的规律，仔细观察后发现其实确实粗略来看是每10个出现一个和为10的倍数

因为对于一个十位数，例如1X，这时候个位只有10 - 1 = 9才满足，对于一个三位数例如12X，则由于1和2是确定的，所以在120 - 129这十个数的范围内（也就是个位0 - 9）只有10 - 1 - 2 = 7，即127满足，对于大一点的数，例如637859X，6 + 3 + 7 + 8 + 5 + 9 = 38，因为最后一位肯定只能为正（不可能为-8），所以只有最后一位为40 - 38 = 2，即6378592符合条件，所有的数都这么看，会发现每次高位确定了，其实在个位上只有0 - 9当中的一个数能够满足，这就印证了num / 10的规律

最后只要注意下右边界就行了，即

0 - 90 -> 90 / 10 = 9个

但是

0 - 91 ->90 / 10 + 1 = 10个（由于91这个边界多加了一个）

处理方法是从90 / 10 * 10 = 90 -->91暴力扫一下，看有没有多出来一个能够满足各位和是10的倍数的，发现91是，所以就在9的基础上加1，因为我们证明了个位0 - 9有且只有一个数满足，所以只要扫过去发现一个直接返回就行了

#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std;

long long get(long long num) {
	long long n = num / 10 * 10;
	bool flag = false;
	for (; n <= num; n++) {
		long long tn = n, sum = 0;
		while (tn) {
			sum += tn % 10;
			tn /= 10;
		}
		if (sum % 10 == 0) {
			flag = true;
			break;
		}
	}
	if (flag)
		return 1;
	else
		return 0;
}

long long solve(long long num) {
	if (num < 0) return 0;
	if (num < 10) return 1;
	return num / 10 + get(num);
}



int main()
{
	int T;
	long long a, b;
	scanf("%d", &T);
	for (int t = 1; t <= T; t++) {
		scanf("%I64d%I64d", &a, &b);
		printf("Case #%d: %I64d\n", t, solve(b) - solve(a - 1));
	}
	return 0;
}