誰可以幫我解釋一下這程序跑法??

[shihyuyao]

void permute(char a[], int k, int input_var)
{
int i,j, inversion=0;
char temp;
if ( k == input_var )
{
printf("(");
for ( i=1; i<input_var; i++ )
printf(" %d,",a[i]);
printf(" %d )",a[i]);

for ( i=1; i<=input_var; i++ )
for ( j = i+1; j<=input_var; j++)
if (a[i] > a[j])
inversion++;

printf(" Inversion = %2d, ",inversion);
if ( (inversion %2) == 1 )
printf("odd permutation.\n");
else
printf("even permutation.\n");
}
else
{
for ( i=k; i<=input_var; i++)
{
temp=a[k]; a[k]=a[i]; a[i]=temp;
permute ( a, k+1, input_var);
temp=a[k]; a[k]=a[i]; a[i]=temp;
}
}
}

int main(int argc, char* argv[])
{
int i, input_var;
char setofnumber[10], ch;

for (i=0; i<10; i++)





permute ( a, k+1, input_var);
temp=a[k]; a[k]=a[i]; a[i]=temp;
}
}
}

int main(int argc, char* argv[])
{
int i, input_var;
char setofnumber[10], ch;

for (i=0; i<10; i++)
setofnumber[i]=i;

printf("Please input the number of entries to permute\n");
scanf("%d",&input_var);
permute(setofnumber, 1, input_var);

return 0;
}


上面是完整程式碼


____________________________________________________________________
下面這程式是怎麼跑的~~可以跟我說明一下

好像是利用遞迴嗎??


void permute(char a[], int k, int input_var)
{
int i,j, inversion=0;
char temp;
if ( k == input_var )
{
printf("(");
for ( i=1; i<input_var; i++ )
printf(" %d,",a[i]);
printf(" %d )",a[i]);

for ( i=1; i<=input_var; i++ )
for ( j = i+1; j<=input_var; j++)
if (a[i] > a[j])
inversion++;

printf(" Inversion = %2d, ",inversion);
if ( (inversion %2) == 1 )
printf("odd permutation.\n");
else
printf("even permutation.\n");
}
else
{
for ( i=k; i<=input_var; i++)
{
temp=a[k]; a[k]=a[i]; a[i]=temp;
permute ( a, k+1, input_var);
temp=a[k]; a[k]=a[i]; a[i]=temp;
}
}
}

[ltc_mouse]

楼主，偶不习惯打繁体字，没关系吧？

/*
Desc: 该函数求出序列a[k...input_var]的全排列,采用递归的方法
Para: a -----序列 a[1...9]
k -----起始位置
input_var ---- 结束位置
*/
void permute(char a[], int k, int input_var)
{
int i,j, inversion=0;
char temp;
if ( k == input_var ) /* 起始位置与结束位置一致，序列只有一个元素，排列结束 */
{
printf("(");
for ( i=1; i<input_var; i++ ) /* 输出一个序列 */
printf(" %d,",a[i]);
printf(" %d )",a[i]);

for ( i=1; i<=input_var; i++ ) /* 计算序列a[1...input_var]的逆序数 */
for ( j = i+1; j<=input_var; j++)
if (a[i] > a[j])
inversion++;

printf(" Inversion = %2d, ",inversion); /* 以下输出序列类型 */
if ( (inversion %2) == 1 )
printf("odd permutation.\n");
else
printf("even permutation.\n");
}
else /* 序列长度大于1时候，递归求解 */
{
for ( i=k; i<=input_var; i++)
{
temp=a[k]; a[k]=a[i]; a[i]=temp; /* a[k]取遍a[k...input_var]中的每个值 */
permute ( a, k+1, input_var); /* 求序列a[k+1...input_var]的全排列 */
temp=a[k]; a[k]=a[i]; a[i]=temp; /* 恢复a[k] */
}
}
}

[llf_hust]

计算全排列

[C_projecting]

怎么会有两个int main(int argc, char* argv[])