PAT 19秋 7-1 Forever (20分)

7-1 Forever (20分)

"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

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.

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

第一反应肯定是直接暴力枚举，但是K的范围大了些。

可以通过打表找规律发现最后两位都是99 ，所以可以只枚举前K-2位，这样数据是刚好的。

因为太久没做过打表题，都忘了还有这个牛逼的技能。

 

#include 
#define rep(i,a,n) for(int i=a;ians;
bool cmp(node a,node b){
    if(a.n!=b.n) return a.n>M;
    reo(cas,1,M){
        cin>>k>>m;
        printf("Case %d\n",cas);
        gao();
        if(ans.size()==0) cout<<"No Solution"<