大数据2班第六次堂测

$block1$
FILE *fp1;
    char str[10000+5][110];
    int i, j , n=0;
    if((fp1=fopen("case1.in","r"))==NULL)
    return 0;
    while((fgets(str[n],110,fp1))!=NULL)
        n++;
        for(i=0;i<n;i++)
        strlwr(str[i]);
    //int temp;
     int flag=1;
    for(i=0;i<n;i++)
    {
         j=2;  
       flag=1;
        while(str[i][j]!='\0')
      {
       if(str[i][j-2]=='b'&&str[i][j-1]=='a'&&str[i][j]=='d')
       {
           flag=0;
           break;
       }
       j++;
      }
      if(!flag)  break;
    }
    if(!flag)  printf("YES\n");
    else printf("NO\n");
    fclose(fp1);



3.

```c
FILE *fp1,*fp2;
    char str[10000+5][110], str1[11];
    int i, j , n=0;
    if((fp1=fopen("case1.in","r"))==NULL)
    return 0;
    if((fp2=fopen("answer.txt","w"))==NULL)
    return 0;
   while((fgets(str[n],110,fp1))!=NULL)
        n++;
    //int temp;
    for(i=0;i<n;i+=2)
    {
       fputs(str[i],fp2);
    }
    //printf("%d\n",n);
    fclose(fp1);
    fclose(fp2);