cbam

摘要： Eddy's research ITime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total ... 阅读全文

posted @ 2015-03-29 22:18 cbam 阅读(46) 评论(0) 推荐(0)

摘要： Eddy's digital RootsTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Tot... 阅读全文

posted @ 2015-03-28 23:55 cbam 阅读(59) 评论(0) 推荐(0)

摘要： Who's in the MiddleTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Tota... 阅读全文

posted @ 2015-03-28 21:47 cbam 阅读(99) 评论(0) 推荐(0)

摘要：最小公倍数Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s... 阅读全文

posted @ 2015-03-28 21:15 cbam 阅读(82) 评论(0) 推荐(0)

摘要：排序Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): ... 阅读全文

posted @ 2015-03-28 21:05 cbam 阅读(88) 评论(0) 推荐(0)

摘要： Ignatius's puzzleTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total ... 阅读全文

posted @ 2015-03-27 23:32 cbam 阅读(49) 评论(0) 推荐(0)

摘要： A hard puzzleTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Subm... 阅读全文

posted @ 2015-03-27 21:52 cbam 阅读(70) 评论(0) 推荐(0)

摘要：一般方法： int pow1(inta,intb){int r=1;while(b--)r*=a;return r;} 二分求幂int pow2(inta,intb){int r=1,base=a;while(b... 阅读全文

posted @ 2015-03-27 21:31 cbam 阅读(86) 评论(0) 推荐(0)

摘要： A+B for Input-Output Practice (VIII) Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K... 阅读全文

posted @ 2015-03-27 20:51 cbam 阅读(80) 评论(0) 推荐(0)

摘要： A+B for Input-Output Practice (IV)Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (J... 阅读全文

posted @ 2015-03-27 20:42 cbam 阅读(100) 评论(0) 推荐(0)