请问这个C语言程序错在哪里？

1个四位数乘以1个一位数等于另1个四位数，组成这三个数的9个数字分别是1~9里的任意数，且不重复。请按顺序输出这三个数：

输出结果应该是：

173846952

196347852

共两组。

下面这段是我编的程序，哪里错了？怎么改正？

void pm(int n)
{int i,a[9];
if(n==0) return;
else for(i=1;i<=9;i++)
{a[n-1]=i;pm[n-1];}

if((a[0]*1000+a[1]*100+a[2]*10+a[3])*a[4]==(a[5]*1000+a[6]*100+a[7]*10+a[8]))
{for(i=0;i<9;i++) printf("%d",a[i]);printf("\n");}}

main()
{pm(9);}

void pm(int n)
{int i,a[9];
if(n==0) return;
else for(i=1;i<=9;i++)
{a[n-1]=i;pm[n-1];}//pm()是函数，不是数组，把[]改成()，不过改完后你这循环套递归，递归套循环,

//想不出来最后是什么结果，建议好好研究重新写个清晰的代码

if((a[0]*1000+a[1]*100+a[2]*10+a[3])*a[4]==(a[5]*1000+a[6]*100+a[7]*10+a[8]))
{for(i=0;i<9;i++) printf("%d",a[i]);printf("\n");}}

main()
{pm(9);}

以下代码输出的结果是正确的：

#include <vector> using namespace std;

#define BITSUM 4

typedef struct Num_struct { int val; bool used; } Num_t;

Num_t num[9]; int x1[4], x2[4], x3; //x1是第一组的4位数，x3是第二个数，x2是结果的4位数 vector<int> v1, v2;//分别存放X1和X2的排列组合结果

int find_unused() { for (int i = 0; i < 9; i++) { if (num[i].used) continue;

return i; }

return -1; }

void check() { int i = 0, j = 0, k = 0, l = 0;

int a = 0;

//第一组4位数排列 v1.clear(); for (i = 0; i < BITSUM; i++) { for (j = 0; j < BITSUM; j++) { if (j == i) continue;

for (k = 0; k < BITSUM; k++) { if (k == j || k == i) continue;

for (l = 0; l < BITSUM; l++) { if (l == i || l == j || l == k) continue;

a = x1[i] * 1000 + x1[j] * 100 + x1[k] * 10 + x1[l]; v1.push_back(a); } } } }

//结果4位数排列 v2.clear(); for (i = 0; i < BITSUM; i++) { for (j = 0; j < BITSUM; j++) { if (j == i) continue;

for (k = 0; k < BITSUM; k++) { if (k == j || k == i) continue;

for (l = 0; l < BITSUM; l++) { if (l == i || l == j || l == k) continue;

a = x2[i] * 1000 + x2[j] * 100 + x2[k] * 10 + x2[l]; v2.push_back(a); } } } }

int v1_size = v1.size(); int v2_size = v2.size();

for (i = 0; i < v1_size; i++) { for (j = 0; j < v1_size; j++) { if (v1[i] * x3 == v2[j])//符合条件、输出结果 { printf("%d * %d = %d\n", v1[i], x3, v2[j]); } } } }

int pm(int start, int layer) { if (layer == BITSUM) { int j = 0, k = 0, n; for ( j = 0; j < 9; j++) { if (num[j].used) continue;

//查找第二个数x3， x3 = num[j].val; num[j].used = true;

//查找第三组的4位数x2， for (k = 0; k < BITSUM; k++) { n = find_unused(); x2[k] = num[n].val; num[n].used = true; } check();//检查是否有符合条件的

num[ x3 - 1].used = false; for (k = 0; k < BITSUM; k++) { num[ x2[k] - 1 ].used = false; } }

return 0; }

int i = start - 1; for (; i >= 0; i--) { if (num[i].used) continue;

//查找第一组的4位数x1， num[i].used = true; x1[layer] = num[i].val;

pm(i, layer+1);//递归查找

num[i].used = false; }

return 0; }

int main(int argc, char* argv[]) { for (int i = 0; i < 9; i++) { num[i].val = i+1; num[i].used = false; }

pm(9, 0);

return 0; }