2019 PAT甲級秋季考試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

暴力王打表發現最後兩位一定是99~

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define INF 0x3f3f3f3f
int gcd(int a,int b)
{
	return a%b==0?b:gcd(b,a%b);
}
bool prime(int x)
{
	if(x<=2) return false;
	for(int i=2;i<=sqrt(x);i++)
	{
		if(x%i==0)
		return false;
	}
	return true;
}
int sumd(int x)
{
	int sum=0;
	while(x)
	{
		sum+=x%10;
		x=x/10;
	}
	return sum;
}
struct sb
{
	int op,val;
}p[1000005];
bool cmp(sb u,sb w)
{
	if(u.op!=w.op)return u.op<w.op;
	else return u.val<w.val;
}
int main()
{
	int t,n,m,k,i,j,o=1;
	scanf("%d",&t);
	while(t--)
	{	
		int flag=0,uio=0;
		scanf("%d%d",&k,&m);
		printf("Case %d\n",o++);
		if(k==4)
		{
			for(i=10;i<=99;i++)
			{
				if(sumd(i)+18==m)
				{
					int op=sumd(i+1);
					if(prime(gcd(m,op)))
					{
						p[uio].op=op;
						p[uio].val=i*100+99;
						uio++;
						//printf("%d %d\n",op,i*100+99);
						flag++;
					}
				}
			}
		}
		else if(k==5)
		{
			for(i=100;i<=999;i++)
			{
				if(sumd(i)+18==m)
				{
					int op=sumd(i+1);
					if(prime(gcd(m,op)))
					{
						p[uio].op=op;
						p[uio].val=i*100+99;
						uio++;
						//printf("%d %d\n",op,i*100+99);
						flag++;
					}
				}
			}
		}
		else if(k==6)
		{
			for(i=1000;i<=9999;i++)
			{
				if(sumd(i)+18==m)
				{
					int op=sumd(i+1);
					if(prime(gcd(m,op)))
					{
						p[uio].op=op;
						p[uio].val=i*100+99;
						uio++;
						//printf("%d %d\n",op,i*100+99);
						flag++;
					}
				}
			}
		}
		else if(k==7)
		{
			for(i=10000;i<=99999;i++)
			{
				if(sumd(i)+18==m)
				{
					int op=sumd(i+1);
					if(prime(gcd(m,op)))
					{
						p[uio].op=op;
						p[uio].val=i*100+99;
						uio++;
						//printf("%d %d\n",op,i*100+99);
						flag++;
					}
				}
			}
		}
		else if(k==8)
		{
			for(i=100000;i<=999999;i++)
			{
				if(sumd(i)+18==m)
				{
					int op=sumd(i+1);
					if(prime(gcd(m,op)))
					{
						p[uio].op=op;
						p[uio].val=i*100+99;
						uio++;
						//printf("%d %d\n",op,i*100+99);
						flag++;
					}
				}
			}
		}
		else if(k==9)
		{
			for(i=1000000;i<=9999999;i++)
			{
				if(sumd(i)+18==m)
				{
					int op=sumd(i+1);
					if(prime(gcd(m,op)))
					{
						p[uio].op=op;
						p[uio].val=i*100+99;
						uio++;
						//printf("%d %d\n",op,i*100+99);
						flag++;
					}
				}
			}
		}
		if(flag==0)
		printf("No Solution\n");
		else
		{
			sort(p,p+uio,cmp);
			for(i=0;i<uio;i++)
			printf("%d %d\n",p[i].op,p[i].val);
		}
	}
}