2021年8月6日
位运算
摘要： &(与)、| (或)、^(异或)、~ (非/取反) 判断奇偶数 x&1=1为奇数 =0为偶数 原因：奇数最后一位是1，&1后为1，偶数最后一位为0，&1后为0 1.将整数的二进制奇偶位互换 package 蓝桥杯算法; import java.util.Scanner; /** * 例如 1001 阅读全文
posted @ 2021-08-06 21:24 guoyuxin3 阅读(58) 评论(0) 推荐(0)
hdu1702 ACboy needs your help again!
摘要： ACboy needs your help again! Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 22399 Accepted Submi 阅读全文
posted @ 2021-08-06 18:34 guoyuxin3 阅读(31) 评论(0) 推荐(0)
hdu1062 Text Reverse
摘要： Text Reverse Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 64691 Accepted Submission(s): 24779 阅读全文
posted @ 2021-08-06 17:48 guoyuxin3 阅读(33) 评论(0) 推荐(0)
hdu4841 圆桌问题
摘要： 圆桌问题 Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 15263 Accepted Submission(s): 5166 Problem D 阅读全文
posted @ 2021-08-06 17:19 guoyuxin3 阅读(28) 评论(0) 推荐(0)