Good Triples

题面翻译

题目描述

给定一个整数 $n$，我们称三元组 $(a,b,c)$ 是“好的”，当且仅当 $a+b+c=n$ 并且 $digsum(a)+digsum(b)+digsum(c)=digsum(n)$。这里的 $digsum(x)$ 指的是 $x$ 各个数位之和。

并且三元组的顺序也很重要。例如 $(4,12,10)$ 和 $(10,12,4)$ 不是相同的三元组。

求不同三元组的数量。

题目描述

Given a non-negative integer number $ n $ ( $ n \ge 0 $ ). Let’s say a triple of non-negative integers $ (a, b, c) $ is good if $ a + b + c = n $ , and $ digsum(a) + digsum(b) + digsum(c) = digsum(n) $ , where $ digsum(x) $ is the sum of digits of number $ x $ .

For example, if $ n = 26 $ , then the pair $ (4, 12, 10) $ is good, because $ 4 + 12 + 10 = 26 $ , and $ (4) + (1 + 2) + (1 + 0) = (2 + 6) $ .

Your task is to find the number of good triples for the given number $ n $ . The order of the numbers in a triple matters. For example, the triples $ (4, 12, 10) $ and $ (10, 12, 4) $ are two different triples.

输入格式

The first line of input contains a single integer $ t $ ( $ 1 \le t \le 10^4 $ ) — the number of test cases. Descriptions of test cases follow.

The first and only line of the test case contains one integer $ n $ ( $ 0 \le n \le 10^7 $ ).

输出格式

For each test case output one integer, the number of good triples for the given integer $ n $ . Order of integers in a triple matters.

样例 #1

样例输入 #1

12
11
0
1
2
3
4
5
3141
999
2718
9999999
10000000

样例输出 #1

9
1
3
6
10
15
21
1350
166375
29160
1522435234375
3

提示

In the first example, the good triples are $ (0, 0, 11) $ , $ (0, 1, 10) $ , $ (0, 10, 1) $ , $ (0, 11, 0) $ , $ (1, 0, 10) $ , $ (1, 10, 0) $ , $ (10, 0, 1) $ , $ (10, 1, 0) $ , $ (11, 0, 0) $ .

In the second example, there is only one good triple $ (0, 0, 0) $ .

因为是3元组，可以算一下0~9每一个数可以有多少种三元方案，每一个数位的三元方案全部乘起来就是总方案数

const int N = 4e5 + 10;
int a[N];
void solve()
{

    int n;
    cin >> n;
    int ans = 1;
    while (n)
    {
        ans *= a[n % 10];
        n /= 10;
    }
    cout << ans << '\n';
}
signed main()
{
    buff;
    int t = 1;
    cin >> t;
    for (int i = 0; i <= 9; i++)
        for (int j = 0; j <= 9; j++)
            for (int k = 0; k <= 9; k++)
                if (i + j + k <= 9) a[i + j + k]++;
    while (t--) solve();
    return 0;
}