开个size为[1000]的散列表嘛~
筛法快速求素数后直接逆序判断并标记有多少个就可以了~

是itoa在闹鬼。修改后的程序如下，仅针对以上谈及错误，其他漏洞未关注：
int main(){
     int i,j,k,l,m,p;
     string str;  //存放4位数字的字符串
     char num[4];
     char val[4];
     for(i = 1000;i <= 9999;i++)
     {
         for(j = 2;j <= i;j++)
         {
             if(i % j == 0)
                 break;
         }
         if(j >= i)
         {   
             //itoa(i,num,10); //有漏洞的操作
             str = to_string(i);
             for(size_t idx=0; idx<str.size(); ++idx){  //遍历字符串，取出各个字符，现在为4个
                 num[idx]=str[idx];                 
             }
             for(l = 0,k = 3;k >= 0;k--,l++)
             {
                 val[l] = num[k];
             }
             p=stoi(val);  //p = atoi(val);
             for(m = 2;m <= p;m++)
             {
                 if(p % m == 0)
                     break;
             }
             if(m >= p)
             {
                 i = atoi(num);  
                 printf("%d ",i);                 
             }
         }   
     }
     printf("\n");
 }

实测结果：

咦，怎么发错地方？

是itoa在闹鬼。修改后的程序如下，仅针对以上谈及错误，其他漏洞未关注：
int main(){
      int i,j,k,l,m,p;
      string str;  //存放4位数字的字符串
     char num[4];
      char val[4];
      for(i = 1000;i <= 9999;i++)
      {
          for(j = 2;j <= i;j++)
          {
              if(i % j == 0)
                  break;
          }
          if(j >= i)
          {   
              //itoa(i,num,10); //有漏洞的操作
             str = to_string(i);
              for(size_t idx=0; idx<str.size(); ++idx){  //遍历字符串，取出各个字符，现在为4个
                 num[idx]=str[idx];                 
              }
              for(l = 0,k = 3;k >= 0;k--,l++)
              {
                  val[l] = num[k];
              }
              p=stoi(val);  //p = atoi(val);
              for(m = 2;m <= p;m++)
              {
                  if(p % m == 0)
                      break;
              }
              if(m >= p)
              {
                  i = atoi(num);  
                  printf("%d ",i);                 
              }
          }   
      }
      printf("\n");
  }

谢谢