C language to achieve 24 point game source code

• 2020-06-19 11:23:52
• OfStack

This article shares the specific code of C language to realize 24 points game for your reference. The specific content is as follows

C language to achieve the classical 24-point algorithm

Change the algorithm implementation to C language and run on the linux server. Also modified to display all results.
Note: if the pass is repeated, such as 4,4,7,7, it will echo the repeat result and cannot be cleaned temporarily.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

const double PRECISION = 1E-6;
#define COUNT 4
const int RESULT = 24;
#define STRLEN 50
double number[COUNT] = {0}; // Here, 1 Need to use double,
char expression[COUNT][STRLEN] = {0}; // Save expression

#define TRUE 1
#define FALSE 0

int cnt = 0;

void Test(int n)
{
int i = 0;
int j = 0;
int len = 0;
// End of the recursive
if(1 == n){
if(number[0] == RESULT)
{
//  Avoid parentheses before and after output
for (i = 1; i < strlen(expression[0]) - 1; i++)
{
printf("%c", expression[0][i]);
}
printf("\n");
cnt++;
return;
}
else
return;
}
// A recursive process
for(i=0;i<n;i++){
for(j=i+1;j<n;j++){
double a,b;
char expa[STRLEN] = {0};
char expb[STRLEN] = {0};
a=number[i];
b=number[j];
//  delete number[j] Elements, number[n-1] fill
number[j]=number[n-1];
strcpy(expa, expression[i]);
strcpy(expb, expression[j]);
//  delete expression[j] Elements, expression[n-1] fill
strcpy(expression[j], expression[n-1]);

len= strlen(expression[i]);
snprintf(expression[i], STRLEN, "(%s+%s)", expa, expb);
number[i]=a+b;
Test(n-1);
// There are two ways to do it ,a-b with b-a
len= strlen(expression[i]);
snprintf(expression[i], STRLEN, "(%s-%s)", expa, expb);
number[i]=a-b;
Test(n-1);
if(a != b)
{
len= strlen(expression[i]);
snprintf(expression[i], STRLEN, "(%s-%s)", expb, expa);
number[i]=b-a;
Test(n-1);
}
//  The multiplication
len= strlen(expression[i]);
snprintf(expression[i], STRLEN, "(%s*%s)", expa, expb);
number[i]=a*b;
Test(n-1);
// There are two ways of dividing ,a/b with b/a
if(b!=0){
len= strlen(expression[i]);
snprintf(expression[i], STRLEN, "(%s/%s)", expa, expb);
number[i]=a/b;
Test(n-1);
}
if((a!=0) && (a != b)){
len= strlen(expression[i]);
snprintf(expression[i], STRLEN, "(%s/%s)", expb, expa);
number[i]=b/a;
Test(n-1);
}
// Return an array
number[i]=a;
number[j]=b;
strcpy(expression[i], expa);
strcpy(expression[j], expb);
}
}
return;
}
int main(int argc, char **argv)
{
int i = 0;

if(5 != argc)
{
printf("arg err\n");
return 0;
}

for(i=0;i<COUNT;i++)
{
char buffer[20];
number[i] = atoi(argv[i + 1]);
strcpy(expression[i], argv[i + 1]);
}

Test(COUNT);

if(0 != cnt)
{
printf("Total[%d], Success\n", cnt);
}
else
{
printf("Fail\n");
}
return 0;
}

The operation results are as follows:

andy@ubuntu14:~/work\$ ./test 5 6 7 8
((5+7)-8)*6
(5+7)*(8-6)
8/((7-5)/6)
(6/(7-5))*8
6/((7-5)/8)
(8/(7-5))*6
(6*8)/(7-5)
((5-8)+7)*6
(7-(8-5))*6
(5+7)*(8-6)
(6*8)/(7-5)
(5+(7-8))*6
(5-(8-7))*6
Total[13], Success
andy@ubuntu14:~/work\$ ./test 7 7 7 7
Fail

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