首页 > 代码库 > nyoj 218 Dinner(水题)
nyoj 218 Dinner(水题)
Dinner
时间限制:100 ms | 内存限制:65535 KB
难度:1
- 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
- 提示
- The tableware only contains: bowl, knife, fork and chopsticks.
- 来源
- 辽宁省10年省赛
- 上传者
ACM_李如兵
南阳OJ上的说是贪心,我也是醉了,太水了,不考虑最后元素的空格竟然也能过,我。。。。。。
代码如下:
#include<stdio.h> #include<string.h> char a[110][110]; int main() { int n,i; while(~scanf("%d",&n)) { for(i=0;i<n;i++) { scanf("%s",a[i]); } for(i=0;i<n;i++) { if(strcmp(a[i],"bowl")==0) printf("bowl "); if(strcmp(a[i],"knife")==0) printf("knife "); if(strcmp(a[i],"fork")==0) printf("fork "); if(strcmp(a[i],"chopsticks")==0) printf("chopsticks "); } printf("\n"); } return 0; }
nyoj 218 Dinner(水题)
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。