Fill a 3x3 grid with 9 of the numbers 1~N (N>0), and fill each grid with 1 integer, so that the sum of the integers in two adjacent grids is a prime number. Seek various number filling methods that meet the above requirements.
[Analysis]
Use heuristics to find the solution of the problem, that is, starting from the first square, find a reasonable integer for the current square to fill in, and after filling in the current position correctly, look for the next square to fill in A reasonable integer. If you can't find a reasonable fillable integer for the current square, you must go back to the previous square and adjust the number of the previous square. When the ninth square is also filled with a reasonable integer, a solution is found, the solution is output, and the ninth filled integer is adjusted to continue searching for the next solution. In order to check the rationality of filling integers in the current square, the two-dimensional array checkMatrix is introduced to store the serial numbers of adjacent squares that need to be checked for rationality.
In order to find a filling method that meets the requirements of 9 numbers, fill in an integer at the current position in a certain order (such as from small to large), and then check whether the currently filled integer can meet the requirements. If the requirements are met, continue to use the same method to fill in integers for the next square. If the recently filled integer does not meet the requirements, change the filled integer. If you have tried all possible integers for the current square, but the requirements are not met, you have to go back to the previous square (backtracking) and adjust the integer filled in the square. Repeat expansion, inspection, and adjustment in this way until a solution that meets the requirements of the problem is found, and the solution is output.
code:
#include<stdio.h>
#define N 12
int b[N + 1];
int a[10];/*存放方格填入的整数*/
int total = 0;/*共有多少种填法*/
int checkmatrix[][3] = { { -1 },{ 0,-1 },{ 1,-1 },
{ 0,-1 },{ 1,3,-1 },{ 2,4,-1 },
{ 3,-1 },{ 4,6,-1 },{ 5,7,-1 } };
void write(int a[])
/*输出方格中的数字*/
{
int i, j;
for (i = 0; i < 3; i++)
{
for (j = 0; j < 3; j++)
printf("%3d", a[3 * i + j]);
printf("\n");
}
}
int isprime(int m)
/*判断m是否是质数*/
{
int i;
int primes[] = { 2,3,5,7,11,17,19,23,29,-1 };
if (m == 1 || m % 2 == 0)
return 0;
for (i = 0; primes[i]>0; i++)
if (m == primes[i])
return 1;
for (i = 3; i*i <= m;)
{
if (m%i == 0)
return 0;
i += 2;
}
return 1;
}
int selectnum(int start)
/*从start开始选择没有使用过的数字*/
{
int j;
for (j = start; j <= N; j++)
if (b[j])
return j;
return 0;
}
int check(int pos)
/*检查填入的pos位置是否合理*/
{
int i, j;
if (pos < 0)
return 0;
/*判断相邻的两个数是否是质数*/
for (i = 0; (j = checkmatrix[pos][i]) >= 0; i++)
if (!isprime(a[pos] + a[j]))
return 0;
return 1;
}
int extend(int pos)
/*为下一个方格找一个还没有使用过的数字*/
{
a[++pos] = selectnum(1);
b[a[pos]] = 0;
return pos;
}
int change(int pos)
/*调整填入的数，为当前方格寻找下一个还没有用到的数*/
{
int j;
/*如果所有数都被使用，则回溯*/
while (pos >= 0 && (j = selectnum(a[pos] + 1)) == 0)
b[a[pos--]] = 1;
if (pos < 0)
return -1;
b[a[pos]] = 1;
a[pos] = j;
b[j] = 0;
return pos;
}
void find()
/*查找*/
{
int ok = 0, pos = 0;
a[pos] = 1;
b[a[pos]] = 0;
do
{
if (ok)
if (pos == 8)
{
total++;
printf("第%d种填法\n", total);
write(a);
pos = change(pos);/*调整*/
}
else
pos = extend(pos);/*扩展*/
else
pos = change(pos);/*调整*/
ok = check(pos);/*检查*/
} while (pos >= 0);
}
void main()
{
int i;
for (i = 1; i <= N; i++)
b[i] = 1;
find();
printf("共有%d种填法\n", total);
getchar();
}result:
