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nyoj 218 Dinner【简单字符串处理】

来源:互联网 作者:佚名 时间:2015-09-01 23:54
Java版本说明: Java开发中,经常遇到从GBK到Unicode码的转换;大家经常做的做法,大概有2种,一种是:native2ascii;只要保存好文件,每次编辑好文件,然后使

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


思路:

      直接遍历所有的字符串,,看看有没有题上给的那四个字符串,如果有的话就输出,如果没有的话,就什么也不输出!但是需要注意输出的格式,在第一个字符串前面没有空格,所以要控制好这一点!

代码:

#include <stdio.h> #include <string.h> int main() { int n; char a[15]; int flag; while(scanf("%d",&n)!=EOF) { flag=0; for(int i=1;i<=n;i++) { getchar(); scanf("%s",a); if(strcmp(a,"bowl")==0) { if(flag) printf(" "); else flag=1; printf("bowl"); } else if(strcmp(a,"knife")==0) { if(flag) printf(" "); else flag=1; printf("knife"); } else if(strcmp(a,"fork")==0) { if(flag) printf(" "); else flag=1; printf("fork"); } else if(strcmp(a,"chopsticks")==0) { if(flag) printf(" "); else flag=1; printf("chopsticks"); } } printf("\n"); } }


 

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