各位大虾，我在编写C程序时遇到一个问题，我想对几组数进行排列组合，比如有三组数：
第一组：1,2,3,4,5,6,7,
第二组：11,12,13,14,15,16,17,
第三组：21,22,23,
现在我想从第一组中取出3个数，从第二组取出3个数，从第三组取出3个数，则第一组有35个结果，第二组有35个结果，第三组有1个结果，
然后再将每一组的各个组合和其它组的组合再组合起来，共计1225个组合结果，然后将这1225个组合结果保存到文件中。以下是我的C源程序，
但是总不能输出正确的结果，运行到后面就会出乱码，哪位好心人可否帮忙指教一下？在此先行谢过！


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

fact(rr)
int rr;
{long f;
  if (rr==0||rr==1) f=1;
     else f=fact(rr-1)*rr;
  return(f);
}

cnm(a,b)
int a,b;
{long c;
  if (a<b) return(0);
     else {c=fact(a)/fact(b)/fact(a-b);return(c);}
}

main()
{  int a[20],r,max_val,j;
   static int seq[20]={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18};
   int nm[20],kk,tnm=1;
  int fenduan,k,i,m,n,lp,mn;
  int data[18][35];
  int geshu[15],cc[15],ge[15];
  int f[8][40][10];
  int df[3276][10];
  int wei[10],xuanshu;
  FILE *fp;

  /*输入分段数及每段的数据，并计出每段的数据个数wei[i]。*/
  printf("Pls input duanshu :");
  scanf("%d",&fenduan);
  for(i=1; i <= fenduan; i++)
     {j=1;k=1;
      while (k!=0)
        {printf("pls input shuju:");
         scanf("%d",&data[i][j]);
           if (data[i][j]==0) k=0; j=j+1;}
      wei[i]=j-2;
}

  /*输出每段的数据*/
  for(i=1; i <= fenduan; i++)
     {printf("Gong %3d ge shu.----",wei[i]);
      printf("Di %d zhu shu: ",i);
      for(j=1;j<=wei[i];j++)
     printf("%3d, ",data[i][j]);
      printf("\n");}

  /*输入每段需要选出的个数geshu[i]，并计出总共选出的个数xuanshu。*/
  xuanshu=0;
  for(i=1;i <= fenduan; i++)
     {printf("Pls input Di %d zhu shuo xu ge shu:",i);
      scanf("%d",&geshu[i]);
      xuanshu=xuanshu+geshu[i];}
      printf("Gong xuan chu %3d ge shu",xuanshu);
      printf("\n");

  /*取每一段的各种组合，并计出每段的组合数nm,各段组合数相乘为总组合数tnm*/
/*fp=fopen("c.txt","w");*/
for(kk=1; kk <= fenduan; kk++)
     {n=wei[kk]; m=geshu[kk];nm[kk]=cnm(n,m); tnm=tnm*nm[kk];nm[0]=cnm(xuanshu, 7);printf("nm[0]= %d. ",nm[0]);/*奇怪的是这个nm[0]总是会等于-5，不知什么原因*/
       for (i=0;i<m;i++)
         a[i]=seq[i];
      for (i=0; i<m; i++)
             {f[kk][1][i+1]=data[kk][a[i]];printf("%3d, ",a[i]);}
      for (k=1;k<nm[kk];k++)
            {
               max_val=seq[n-1];
               r=m-1;
               while(a[r]==max_val)
                        {
            r-=1;
            max_val-=1;
                         }
                a[r]+=1;
                for( j=r+1;j<m;j++)
                      a[j]=a[j-1]+1;
                for (i=0; i<m; i++)
                       f[kk][k+1][i+1]=data[kk][a[i]];
              }
     for (i=1;i<=nm[kk];i++)
         {printf("%3d: ",i);/*fprintf(fp,"%3d: ",i);*/
          for (j=1;j<=m;j++)
              {printf("%3d, ",f[kk][i][j]);fprintf(fp,"%3d, ",f[kk][i][j]);}
          printf("\n");/*fprintf(fp,"\n");*/
      }
  }/*每段组合运行结束*/

/*各段合成总组合*/
for (i=1; i<=fenduan; i++)  {cc[i]=1; ge[i]=0;}
for (i=1; i<fenduan; i++)
       for (j=i+1; j<=fenduan; j++)
             cc[i]=cc[i]*nm[j];
for (i=2; i<=fenduan; i++)
       for (j=1; j<i; j++)
             ge[i]=ge[i]+geshu[j];

geshu[0]=0;k=1;r=1;cc[0]=0;
for(i=1; i<=fenduan; i++)
      { lp=0;
        en2: for (r=1; r<=nm[i]; r++)
             for (kk=(cc[i]*(r-1+lp)+1); kk<=(cc[i]*(r+lp)); kk++)
                  {k=1;
                    for (j=(ge[i]+1); j<=(ge[i]+geshu[i]); j++)
                             {df[kk][j]=f[i][r][k];k=k+1;  }
                   }
         if (kk<tnm) {lp=lp+nm[i]; goto en2;}
       }         
/*排序*/
kk=1;
while (kk<=tnm)
         {for (i=1; i<xuanshu; i++)
                {k=i;
                  for (j=i+1; j<=xuanshu; j++)
                         if (df[kk][k]>df[kk][j])
                             k=j;
                  if (k!=i)
                      {r=df[kk][i]; df[kk][i]=df[kk][k]; df[kk][k]=r;}
                  }
            kk++;
           }

/*输出结果 */
fp=fopen("c.txt","w");
for (kk=1;kk<=tnm;kk++)
       {printf("%6d: ",kk);fprintf(fp,"%6d: ",kk);
         for (k=1;k<=xuanshu;k++)
               {printf("%3d, ",df[kk][k]);fprintf(fp,"%3d",df[kk][k]);}
     printf("\n");fprintf(fp,"\n");
        }


fclose (fp);

}
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