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摘要: Max Sum Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 206582 Accepted Submission(s): 48294 Prob 阅读全文
posted @ 2016-04-23 19:13 Ritchie丶 阅读(219) 评论(0) 推荐(0) 编辑
摘要: The area Problem Description Ignatius bought a land last week, but he didn't know the area of the land because the land is enclosed by a parabola and 阅读全文
posted @ 2016-04-23 19:09 Ritchie丶 阅读(276) 评论(0) 推荐(0) 编辑
摘要: Prison Transfer Time Limit: 1000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u Submit Status Description The prison of your city has n priso 阅读全文
posted @ 2016-04-23 18:55 Ritchie丶 阅读(189) 评论(0) 推荐(0) 编辑
摘要: Police Recruits Time Limit: 1000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u Submit Status Description The police department of your city 阅读全文
posted @ 2016-04-23 18:51 Ritchie丶 阅读(309) 评论(0) 推荐(0) 编辑
摘要: Sereja and Mirroring Time Limit: 1000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u Submit Status Description Let's assume that we are given 阅读全文
posted @ 2016-04-23 18:46 Ritchie丶 阅读(223) 评论(0) 推荐(0) 编辑
摘要: Sereja and Mugs Time Limit: 1000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u Submit Status Description Sereja showed an interesting game t 阅读全文
posted @ 2016-04-23 18:44 Ritchie丶 阅读(252) 评论(0) 推荐(0) 编辑
摘要: Magic Formulas Time Limit: 2000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u Submit Status Description People in the Tomskaya region like m 阅读全文
posted @ 2016-04-23 18:40 Ritchie丶 阅读(298) 评论(0) 推荐(0) 编辑
摘要: Megacity Time Limit: 2000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u Submit Status Description The administration of the Tomsk Region fir 阅读全文
posted @ 2016-04-23 18:37 Ritchie丶 阅读(240) 评论(0) 推荐(0) 编辑
摘要: Squats Time Limit: 1000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u Submit Status Description Pasha has many hamsters and he makes them wo 阅读全文
posted @ 2016-04-23 18:36 Ritchie丶 阅读(325) 评论(0) 推荐(0) 编辑
摘要: 描述:关于辗转相除法的具体实现在这里就不具体说明了,本文要记录的是辗转相除法应用于求最大公约数的算法证明过程。 假设: 求m和n的最大公约数。a,b分别是m除以n的商和余数,即m=na+b。gcd(m,n)表示m和n的最大公约数。 求证:gcd(m,n)=gcd(n,b) 证明: 设c=gcd(m, 阅读全文
posted @ 2016-04-23 18:11 Ritchie丶 阅读(461) 评论(0) 推荐(0) 编辑
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