"Forever number" is a positive integer A with K digits, satisfying the following constrains:

the sum of all the digits of A is m;
the sum of all the digits of A+1 is n; and
the greatest common divisor of m and 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). Then N lines follow, each gives a pair of K (3<K<10) and m (1<m<90), of which the meanings are given in the problem description.

Output Specification:

For each pair of K and m, first print in a line Case X, where X is the case index (starts from 1). Then print n and 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. If still not unique, output in the ascending order of A. If there is no solution, output No Solution.

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

#include<iostream>
#include<cmath>
#include<vector>
#include<algorithm>

using namespace std;
typedef struct {
int n, num;
}number;

int gcd(int n, int m){
return m == 0 ? n : gcd(m, n%m);
}
bool isPrime(int n) {
if (n < 3) return false;
int sqrtn = (int)sqrt(n);
for (int i = 2; i <= sqrt(n); i++) {
if (n%i == 0) return false;
}
return true;
}
int getSum(int n) {
return n == 0 ? 0 : getSum(n / 10) + n % 10;
}

int main() {
int num, k, m;
scanf("%d", &num);
for (int i = 0; i < num; i++) {
vector<number> ans;
scanf("%d %d", &k, &m);
printf("Case %d\n", i + 1);
int minn = (int)pow(10, k - 3);
int maxn = minn * 10;
for (int j = minn; j < maxn; j++) {
int tmp = j * 100 + 99;
int sum1 = getSum(tmp + 1);
if (getSum(tmp) == m && isPrime(gcd(sum1, m)))
ans.push_back({ sum1, tmp });
}
if(ans.empty()) printf("No Solution\n");
else {
sort(ans.begin(), ans.end(), [](number n1, number n2) {
return n1.n == n2.n ? n1.num < n2.num : n1.n < n2.n;
});
for (auto it : ans) printf("%d %d\n", it.n, it.num);
}
}
return 0;
}