1932: Dinner
时间限制: 100 Sec 内存限制: 64 MB
题目描述
Little A is one member of ACM team. He had just won the gold in World Final. To celebrate, he decided to invite all to have one meal. As bowl, knife and other tableware is not enough in the kitchen, Little A goes to take backup tableware in warehouse. There are many boxes in warehouse, one box contains only one thing, and each box is marked by the name of things inside it. For example, if "basketball" is written on the box, which means the box contains only basketball. With these marks, Little A wants to find out the tableware easily. So, the problem for you is to help him, find out all the tableware from all boxes in the warehouse.
输入
There are many test cases. Each case contains one line, and one integer N at the first, N indicates that there are N boxes in the warehouse. Then N strings follow, each string is one name written on the box.
输出
For each test of the input, output all the name of tableware.
样例输入
3 basketball fork chopsticks
2 bowl letter样例输出
fork chopsticks
bowl代码一：
#include<stdio.h>
#include<string.h>
#include<stdlib.h>//比较两个字符串。设这两个字符串为str1，str2，若str1==str2，则返回零；若str1>str2，则返回正数；若str1<str2，则返回负数。
当且仅当str和str2相等时，结果为0
!strcmp(str,str2)是相当于结果不为0，所以就是两个字符串不相等的意思
当两个字符串不相等时if成立。
char str[100][100];
int main()
{
int n,i;
while(scanf("%d",&n)!=EOF)
{
for(i=0;i<n;i++)
scanf("%s",str[i]);
int k=0;
for(i=0;i<n;i++)
{
if(strcmp(str[i],"bowl")==0||strcmp(str[i],"knife")==0||strcmp(str[i],"fork")==0||strcmp(str[i],"chopsticks")==0)
{
if(k==0)
printf("%s",str[i]);
else printf(" %s",str[i]);
k++;
}
}
printf("\n");
}
return 0;
} 代码二：
#include<stdio.h>
#include<string.h>
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
getchar();
char a[n][100];
char b[]="chopsticks";
char c[]="fork";
char d[]="bowl";
int len[n];
int m[n];
for(int i=0;i<n;i++)
{
m[i]=0;
}
for(int i=0;i<n;i++)
{
scanf("%s",&a[i]);
}
for(int i=0;i<n;i++)
len[i]=strlen(a[i]);
for(int i=0;i<n;i++)
{
for(int j=0;j<len[i];j++)
{
if(a[i][j]==b[j]||a[i][j]==c[j]||a[i][j]==d[j])
m[i]++;
else break;
}
}
for(int i=0;i<n;i++)
{
if(len[i]==m[i])
printf("%s ",a[i]);
}
printf("\n");
}
return 0;
}