2015年8月6日

HDU_1207_汉诺塔2

摘要: 注意long long的范围 阅读全文

posted @ 2015-08-06 11:37 JASONlee3 阅读(325) 评论(0) 推荐(0) 编辑

2015年5月30日

POJ_1611_The Suspect

摘要: The SuspectsTime Limit:1000MSMemory Limit:20000KTotal Submissions:25149Accepted:12329DescriptionSevere acute respiratory syndrome (SARS), an atypical ... 阅读全文

posted @ 2015-05-30 12:29 JASONlee3 阅读(153) 评论(0) 推荐(0) 编辑

POJ_1847_Tram

摘要: TramTime Limit:1000MSMemory Limit:30000KTotal Submissions:11159Accepted:4089DescriptionTram network in Zagreb consists of a number of intersections an... 阅读全文

posted @ 2015-05-30 12:27 JASONlee3 阅读(216) 评论(0) 推荐(0) 编辑

2015年5月27日

POJ_2255_Tree Recovery

摘要: Tree RecoveryTime Limit:1000MSMemory Limit:65536KTotal Submissions:12342Accepted:7712DescriptionLittle Valentine liked playing with binary trees very ... 阅读全文

posted @ 2015-05-27 20:18 JASONlee3 阅读(286) 评论(0) 推荐(0) 编辑

2015年5月26日

Queries for Number of Palindromes(求任意子列的回文数)

摘要: H. Queries for Number of Palindromestime limit per test5 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou've got a... 阅读全文

posted @ 2015-05-26 17:32 JASONlee3 阅读(260) 评论(0) 推荐(0) 编辑

POJ_1163_The triangle

摘要: The TriangleTime Limit:1000MSMemory Limit:10000KTotal Submissions:40079Accepted:24144Description73 88 1 02 7 4 44 5 2 6 5(Figure 1... 阅读全文

posted @ 2015-05-26 17:23 JASONlee3 阅读(171) 评论(0) 推荐(0) 编辑

POJ_3278_Catch That Cow

摘要: Catch That CowTime Limit:2000MSMemory Limit:65536KTotal Submissions:54911Accepted:17176DescriptionFarmer John has been informed of the location of a f... 阅读全文

posted @ 2015-05-26 17:15 JASONlee3 阅读(155) 评论(0) 推荐(0) 编辑

HDU_1166_敌兵布阵

摘要: 敌兵布阵Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 54389Accepted Submission(s): 22819Problem Desc... 阅读全文

posted @ 2015-05-26 15:01 JASONlee3 阅读(217) 评论(0) 推荐(0) 编辑

HDU_1879_继续畅通工程

摘要: 继续畅通工程Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 16654Accepted Submission(s): 7157Problem Des... 阅读全文

posted @ 2015-05-26 14:48 JASONlee3 阅读(158) 评论(0) 推荐(0) 编辑

2015年5月12日

大数减法

摘要: 大数减法void Move(char a[]) //当结果第一位为零时,用此函数将有效位向前移动一位{ int i,len,cnt=0; len=strlen(a); if(a[0]=='0') for(i=0;i0) //当减数的长度大于0,从数组最后... 阅读全文

posted @ 2015-05-12 12:01 JASONlee3 阅读(170) 评论(0) 推荐(0) 编辑

导航