poj 2013 Symmetric Order
2013-10-04 11:12
253 查看
今天早上水过的第3题,简单的字符串处理,其实我只用了输出处理,连排序都没用。。。
#include <stdio.h> int main() { int n; int i; char string[20][30]; int count=0; while(scanf("%d",&n)) { if(n==0) break; for(i=1;i<=n;i++) { scanf("%s",string[i]); //printf("%s\n",string[i]); } printf("SET %d\n",++count); for(i=1;i<=n;i+=2) printf("%s\n",string[i]); if(n%2!=0) n--; for(i=n;i>=2;i-=2) printf("%s\n",string[i]); } return 0; }
相关文章推荐
- poj 2013 Symmetric Order
- poj 2013 Symmetric Order
- (简单递归3.3.1)POJ 2013 Symmetric Order(对称输出)
- poj 2013 Symmetric Order
- poj 2013 Symmetric Order
- poj 2013 Symmetric Order
- OpenJudge/Poj 2013 Symmetric Order
- poj 2013 Symmetric Order
- POJ 2013 Symmetric Order(水~)
- POJ 2013 Symmetric Order (水题)
- POJ 2013 Symmetric Order
- POJ 2013 Symmetric Order 水
- POJ 2013 Symmetric Order(我的水题之路——奇偶输出)
- poj 2013 Symmetric Order 解题报告
- poj 2013 Symmetric Order
- POJ 2013 Symmetric Order
- POJ 2013 - Symmetric Order
- POJ-2013 Symmetric Order-对称排列人名
- OpenJudge/Poj 2013 Symmetric Order
- POJ2013浅析------Symmetric Order