4个回答

((a ⊙ b) ⊙ (c ⊙ d))，符号表示运算符
#include
#include
double num[4];
bool dfs(int n)
{
if(n==1)
{
//n=1时 最终的结果保存在num[0]
if(fabs(num[0]-24)<0.000001)
return true;
else
return false;
}
for(int i=0;i<n;i++)
{
for(int j=i+1;j<n;j++)
{
double x=num[i];
double y=num[j];
//已经使用过的数 使它等于最后一个数
num[j]=num[n-1];
num[i]=x+y;if(dfs(n-1)) return true;
num[i]=x-y;if(dfs(n-1)) return true;
num[i]=y-x;if(dfs(n-1)) return true;
num[i]=x*y;if(dfs(n-1)) return true;
//避免除0
if(y)
{
num[i]=x/y;
if(dfs(n-1)) return true;
}
//避免除0
if(x)
{
num[i]=y/x;
if(dfs(n-1)) return true;
}
//回溯
num[i]=x;
num[j]=y;
}
}
return false;
}
int main()
{
int n;
scanf("%d",&n);
while(n--)
{
scanf("%lf %lf %lf %lf",&num[0],&num[1],&num[2],&num[3]);
if(dfs(4))
puts("Yes");
else
puts("No");
}
return 0;
}

`````` * 24.c 计算4个1到13之间的数（包含）是否能够通过加减乘除达到24.*/
#include <stdio.h>

#define MASK 0xFFFF
#define SHIFT 16
enum {FAILURE = 0, SUCCESS};
const char ops[] = "+*-/";

int test (const int nums[], char *result);
int calculate (int num1, int num2, char op);

int main(int argc, char * argv[])
{
int len = 0;
int nums[4] = {0};
int i;
char *result;
if (argc != 5)
{
printf("Usage: 24.exe num1 num2 num3 num4\n");
exit(1);
}
for (i = 0 ;i < 4 ;i++ )
{
len += strlen (argv[i+1]);
nums[i] = atoi (argv[i+1]);
if (nums[i] < 1 || nums[i] > 13)
{
printf ("所有的数字都应该是1到13正整数\n");
exit(2);
}
}
/*2 pairs of paren, 3 operators, 1 for NULL. */
len += 4 + 3 + 1;
result = (char *)malloc (len);

if (test(nums, result))
printf ("%s\n", result);
else
printf ("No Matched.\n");

if (result != NULL)
free (result);

return 0;
}
/* 测试所有可能的情况，找到一种解法*/
int test (const int nums[], char * result)
{
int I, J, M, N;
int r, s, t;
int ret, ret1, ret2;

/*first loop: I J r M N s t ==> (I r J) t (M s N) */
for (I = 0; I < 4; I++)
{
for (J = 0; J < 4; J++)
{
if (J == I)
continue;
for (r = 0; r < 4; r++)
{
ret1 = calculate (nums[I], nums[J], ops[r]);
if (ret1 <= 0)
continue;
for (M = 0; M < 4; M++)
{
if (M == I || M == J)
continue;
for (N = 0; N < 4; N++)
{
if (N == I || N == J || N == M)
continue;
for (s = 0; s < 4; s++)
{
ret2 = calculate (nums[M], nums[N], ops[s]);
if (ret2 <= 0)
continue;
for (t = 0; t < 4; t++)
{
ret = calculate (ret1, ret2, ops[t]);
if (((ret&MASK)==24) && ((ret>>SHIFT)==0))
{
sprintf (result, "(%d%c%d)%c(%d%c%d)",
nums[I], ops[r], nums[J], ops[t],
nums[M], ops[s], nums[N]);
return SUCCESS;
}
}
}
}
/* second loop: I J r M s N t ==> ((I r J) s M) t N */
for (s = 0; s < 4; s++)
{
ret2 = calculate (ret1, nums[M], ops[s]);
if (ret2 <= 0)
continue;
for (N = 0; N < 4; N++)
{
if (N == I || N == J || N == M)
continue;
for (t = 0; t < 4; t++)
{
ret = calculate (ret2, nums[N], ops[t]);
if (((ret&MASK)==24) && ((ret>>SHIFT)==0))
{
sprintf (result, "((%d%c%d)%c%d)%c%d",
nums[I], ops[r], nums[J], ops[s],
nums[M], ops[t], nums[N]);
return SUCCESS;
}
}
}
}
}
}
}
}
/* third loop: I J M r s N t ==> (I s (J r M)) t N */
for (I = 0; I < 4; I++)
{
for (J = 0; J < 4; J++)
{
if (J == I)
continue;
for (M = 0; M < 4; M++)
{
if (M == I || M == J)
continue;
for (r = 0; r < 4; r++)
{
ret1 = calculate (nums[J], nums[M], ops[r]);
if (ret1 <= 0)
continue;
for (s = 0; s < 4; s++)
{
ret2 = calculate (nums[I], ret1, ops[s]);
if (ret2 <= 0)
continue;
for (N = 0; N < 4; N++)
{
if (N == I || N == J || N == M)
continue;
for (t = 0; t < 4; t++)
{
ret = calculate (ret2, nums[N], ops[t]);
if (((ret&MASK)==24) && ((ret>>SHIFT)==0))
{
sprintf (result, "(%d%c(%d%c%d))%c%d",
nums[I], ops[s], nums[J], ops[r],
nums[M], ops[t], nums[N]);
return SUCCESS;
}
}
}
}
}
}
}
}
/* forth loop: I J M N r s t ==> I t (J s (M r N)) */
for (I = 0; I < 4; I++)
{
for (J = 0; J < 4; J++)
{
if (J == I)
continue;
for (M = 0; M < 4; M++)
{
if (M == I || M == J)
continue;
for (N = 0; N < 4; N++)
{
if (N == I || N == J || N == M)
continue;
for (r = 0; r < 4; r++)
{
ret1 = calculate (nums[M], nums[N], ops[r]);
if (ret1 <= 0)
continue;
for (s = 0; s < 4; s++)
{
ret2 = calculate (nums[J], ret1, ops[s]);
if (ret2 <= 0)
continue;
for (t = 0; t < 4; t++)
{
ret = calculate (nums[I], ret2, ops[t]);
if (((ret&MASK)==24) && ((ret>>SHIFT)==0))
{
sprintf (result, "%d%c(%d%c(%d%c%d))",
nums[I], ops[t], nums[J], ops[s],
nums[M], ops[r], nums[N]);
return SUCCESS;
}
}
}
}
}
}
}
}

return FAILURE;
}
/* 计算两个特定的数和操作符的结果*/
int calculate (int num1, int num2, char op)
{
int numerator_num1, numerator_num2;
int denominator_num1, denominator_num2;
int ret = 0;
int denominator, numerator;

denominator_num1 = num1 >> SHIFT;           //分母
denominator_num2 = num2 >> SHIFT;
numerator_num1 = num1 & MASK;           //分子
numerator_num2 = num2 & MASK;
/* 确定分母，将分子同一化*/
if (denominator_num1 > 0 && denominator_num2 > 0)
{
denominator    = denominator_num1 * denominator_num2;
numerator_num1 = denominator_num2 * numerator_num1;
numerator_num2 = denominator_num1 * numerator_num2;
}
else if (denominator_num1 > 0 && denominator_num2 == 0)
{
denominator    = denominator_num1;
numerator_num2 = denominator_num1 * numerator_num2;
}
else if (denominator_num1 == 0 && denominator_num2 > 0)
{
denominator    = denominator_num2;
numerator_num1 = denominator_num2 * numerator_num1;
}
else
{
denominator = 0;
}
/* 计算*/
if (op == '+')
{
numerator = numerator_num1 + numerator_num2;
}
else if (op == '-')
{
numerator = numerator_num1 - numerator_num2;
}
else if (op == '*')
{
numerator    = numerator_num1 * numerator_num2;
denominator *= denominator;
}
else if (op == '/')
{
if (numerator_num2 > 0 && numerator_num1%numerator_num2 == 0)
{
/* 分子可以整除，分母就没有必要了。*/
numerator   = numerator_num1 / numerator_num2;
denominator = 0;
}
else
{
numerator   = numerator_num1;
denominator = numerator_num2;
}
}
if (denominator>0 && numerator%denominator == 0)
{
numerator   = numerator / denominator;
denominator = 0;
}
ret = (numerator<=0)?numerator:((numerator&MASK) | (denominator<<SHIFT));
return ret;
}

``````
baidu_35042802 好长
4 年多之前 回复

C语言的
#include

void main()
{
int sum,a,b,c,d;
scanf("%d%d%d%d",&a,&b,&c,&d);
sum=a+b+c+d;
if(sum==24)
printf("yes");
else
printf("no");
}