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<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.
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
思路：
枚舉每一位，考慮每一位取值後位數和與m的關係，
註定取不到m的便剪枝：
1. 若當前位數和爲sum，當前位取i,剩餘rest_K - 1位全取9仍然達不到m，則剪枝 。
即 if( sum + i + (rest_K-1) * 9 < m) return ;
2. 若當前位數和位爲sum,當前位取i，已經超過m，則剪枝
即 if ( sum + i > m) return ;
並需要注意首位只能取1 ~ 9，剩下k-1位可取0 ~ 9 。
Case:
#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
struct node{
int n,A;
};
int n,m,N,k;
vector<node> ans;
bool cmp(node a,node b){
if(a.n != b.n) return a.n < b.n;
else return a.A < b.A;
}
bool isPrime(int x){
if(x <= 2) return false;
for(int i = 2; i * i<= x; i++)
if(x % i == 0) return false;
return true;
}
int gcd(int a,int b){
return b == 0 ? a : gcd(b,a%b);
}
int digit_sum(int A){
int sum = 0;
while(A){
sum += A%10;
A/=10;
}
return sum;
}
void DFS(int A,int sum,int rest_K){
if(rest_K == 0){
if(sum == m){
int n = digit_sum(A+1);
if(isPrime(gcd(m,n)))
ans.push_back({n,A});
}
}
else if(rest_K > 0){
for(int i = 0; i <= 9; i++){
if(sum + i + (rest_K-1)*9 >= m && sum + i <= m)
DFS(A*10 + i,sum+i,rest_K-1);
}
}
}
int main(){
scanf("%d",&N);
for(int x = 1; x <= N; x++){
ans.clear();
printf("Case %d\n",x);
scanf("%d%d",&k,&m);
for(int i = 1; i <= 9; i++)
DFS(i,i,k-1);
if(ans.empty()){
printf("No Solution\n");
continue;
}
sort(ans.begin(),ans.end(),cmp);
for(int i = 0; i < ans.size(); i++){
printf("%d %d\n",ans[i].n,ans[i].A);
}
}
return 0;
}