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“Forever number” is a positive integer
A
A
A with
K
K
K digits, satisfying the following constrains:
- the sum of all the digits of
A
A
A is
m
m
m;
- the sum of all the digits of
A
+
1
A+1
A+1 is
n
n
n; and
- the greatest common divisor of
m
m
m and
n
n
n is a prime number which is greater than 2.
Now you are supposed to find these forever numbers.
Input Specification:
Each input file contains one test case. For each test case, the first line contains a positive integer
N
(
≤
5
)
N (≤5)
N(≤5). Then
N
N
N lines follow, each gives a pair of
K
(
3
<
K
<
10
)
K (3<K<10)
K(3<K<10) and
m
(
1
<
m
<
90
)
m (1<m<90)
m(1<m<90), of which the meanings are given in the problem description.
Output Specification:
For each pair of
K
K
K and
m
m
m, first print in a line Case X, where X is the case index (starts from 1). Then print
n
n
n and
A
A
A in the following line. The numbers must be separated by a space. If the solution is not unique, output in the ascending order of
n
n
n. If still not unique, output in the ascending order of
A
A
A. If there is no solution, output No Solution.
Sample Input:
2
6 45
7 80
Sample Output:
Case 1
10 189999
10 279999
10 369999
10 459999
10 549999
10 639999
10 729999
10 819999
10 909999
Case 2
No Solution
Caution:
这道题属实把我愣住了，明明一下子就看出来暴力枚举肯定不行但是就是没想到DFS，愣了一会就去做后面三个题了，后面拐回来还是脑子短路用了暴力，结束后看到的题解主要是用DFS，但是还看到一个根据数学规律来做的，各放一个链接👇
DFS、数学
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