$block1$
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);
大数据2班第六次堂测
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转载自blog.csdn.net/qq_45492531/article/details/103353308
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