摘要: 数组void build(char *s){ int rt=0; for(int i=0;s[i];i++) { int x=s[i]-'a'; if(... 阅读全文
posted @ 2018-08-21 21:36 aeipyuan 阅读(198) 评论(0) 推荐(0) 编辑
摘要: What Are You Talking AboutTime Limit: 10000/5000 MS (Java/Others) Memory Limit: 102400/204800 K (Java/Others)Total Submission(s): 2... 阅读全文
posted @ 2018-08-21 21:23 aeipyuan 阅读(183) 评论(0) 推荐(0) 编辑
摘要: What Are You Talking AboutTime Limit: 10000/5000 MS (Java/Others) Memory Limit: 102400/204800 K (Java/Others)Total Submission(s): 2... 阅读全文
posted @ 2018-08-21 16:00 aeipyuan 阅读(132) 评论(0) 推荐(0) 编辑