第八届省赛 B:Quadrat （打表找规律）

Description

It is well-known that for any n there are exactly four n-digit numbers (including ones with leading zeros) that are self-squares: the last ndigits of the square of such number are equal to the number itself. These four numbers are always suffixes of these four infinite sequences:

...0000000000 
...0000000001 
...8212890625 
...1787109376

For example,   09376²  =87909376, which ends with 09376.

You are required to count the numbers that are almost self-squares: such that each of the last n digits of their square is at most d away from the corresponding digit of the number itself. Note that we consider digits 0 and 9 to be adjacent, so for example digits that are at most 3 away from digit 8 are 5, 6, 7, 8, 9, 0 and 1.

Input

The first line contains the number of test cases t,1≤t≤72. Each of the next t lines contains one test case: two numbers n(1≤n≤ 18) and d(0≤ d≤3).

Output

For each test case, output the number of almost self-squares with length n and the (circular) distance in each digit from the square at most d in a line by itself.

Sample Input

2
5 0
2 1

Sample Output

4
12

Hint

In the second case, number 12's almost self-squares are: 00, 01, 10, 11, 15, 25, 35, 66, 76, 86, 90, 91

题意：

求满足以下条件

①.这个数为n位(可以有前导零)

②.取它的平方的后n位，与它本身每一位对应之差≤d(这里的差指的是数字之间的距离，而这个距离是将数字按圈排列，0与9相邻所求得的)

的数字的个数。

题解：

打表吧

规律就是 res[i][j] = res[i-1][j]*(2*j+1)

代码：

#include <algorithm>
#include <bitset>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <deque>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <vector>
using namespace std;
#define is_lower(c) (c >= 'a' && c <= 'z')
#define is_upper(c) (c >= 'A' && c <= 'Z')
#define is_alpha(c) (is_lower(c) || is_upper(c))
#define is_digit(c) (c >= '0' && c <= '9')
#define min(a, b) ((a) < (b) ? (a) : (b))
#define max(a, b) ((a) > (b) ? (a) : (b))
#define IO                 \
  ios::sync_with_stdio(); \
  cin.tie();              \
  cout.tie();
#define For(i, a, b) for (int i = a; i <= b; i++)
typedef long long ll;
typedef unsigned long long ull;
const ll inf = 0x3f3f3f3f;
;
const ll inf_ll = (ll)1e18;
const ll mod = 1000000007LL;
;
ll res[][];
void init() {
    res[][] = res[][] = ;
    res[][] = res[][] = ;
    ; i < ; i++) {
        ;j <= ; j++)
            res[i][j] = res[i-][j] * ( * j + );
    }
}
int main() {
    init();
    int T;
    cin >> T;
    while(T--){
        int n,d;
        cin>>n>>d;
        cout << res[n][d] << endl;
    }
    ;
}