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ACM POJ 2109 Power of Cryptography

摘要： http://poj.org/problem?id=2109Power of CryptographyTime Limit: 1000MSMemory Limit: 30000KTotal Submissions: 10904Accepted: 5626DescriptionCurrent work in cryptography involves (among other things) large prime numbers and computing powers of numbers among these primes. Work in this area has resulted 阅读全文

posted @ 2011-08-01 20:10 kuangbin 阅读(333) 评论(0) 推荐(0) 编辑

ACM HDU 1070 Milk

摘要： MilkTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5689Accepted Submission(s): 1291Problem DescriptionIgnatius drinks milk everyday, now he is in the supermarket and he wants to choose a bottle of milk. There are many kinds of milk in the supermar 阅读全文

posted @ 2011-08-01 19:19 kuangbin 阅读(771) 评论(0) 推荐(0) 编辑

ACM HDU 1106 排序 (简单题）

摘要：排序Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 15676Accepted Submission(s): 4032Problem Description输入一行数字，如果我们把这行数字中的‘5’都看成空格，那么就得到一行用空格分割的若干非负整数（可能有些整数以‘0’开头，这些头部的‘0’应该被忽略掉，除非这个整数就是由若干个‘0’组成的，这时这个整数就是0）。你的任务是：对这些分割得到的整数，依从小到大的顺序排序输出。Input输入包含多组阅读全文

posted @ 2011-08-01 19:16 kuangbin 阅读(2712) 评论(2) 推荐(0) 编辑

ACM HDU 1108 最小公倍数(完全的水题）

摘要：最小公倍数Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 14564Accepted Submission(s): 7651Problem Description给定两个正整数，计算这两个数的最小公倍数。Input输入包含多组测试数据，每组只有一行，包括两个不大于1000的正整数.Output对于每个测试用例，给出这两个数的最小公倍数，每个实例输出一行。Sample Input10 14Sample Output70SourcePOJ Reco 阅读全文

posted @ 2011-08-01 19:14 kuangbin 阅读(1563) 评论(0) 推荐(0) 编辑

ACM HDU 1157 Who's in the Middle （完全的水题，排序找中值的）

摘要： Who's in the MiddleTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3360Accepted Submission(s): 1646Problem DescriptionFJ is surveying his herd to find the most average cow. He wants to know how much milk this 'median' cow gives: half of 阅读全文

posted @ 2011-08-01 19:04 kuangbin 阅读(2457) 评论(0) 推荐(0) 编辑

ACM HDU 1163 Eddy's digital Roots(数根，找规律）

摘要： Eddy's digital RootsTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1609Accepted Submission(s): 947Problem DescriptionThe digital root of a positive integer is found by summing the digits of the integer. If the resulting value is a single digit 阅读全文

posted @ 2011-08-01 19:01 kuangbin 阅读(448) 评论(0) 推荐(0) 编辑

ACM HDU 1164 Eddy's research I(素数分解，水题）

摘要： Eddy's research ITime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2443Accepted Submission(s): 1492Problem DescriptionEddy's interest is very extensive, recently he is interested in prime number. Eddy discover the all number owned can be divi 阅读全文

posted @ 2011-08-01 18:55 kuangbin 阅读(429) 评论(0) 推荐(0) 编辑

ACM HDU 1170 Balloon Comes! （完全的水题，加减乘除）

摘要： Balloon Comes!Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 10546Accepted Submission(s): 3623Problem DescriptionThe contest starts now! How excited it is to see balloons floating around. You, one of the best programmers in HDU, can get a very bea 阅读全文

posted @ 2011-08-01 18:49 kuangbin 阅读(3577) 评论(0) 推荐(0) 编辑

【转】2011 Multi-University Training Contest 3 - Host by BIT

摘要： T1:Inverting Cups题意是给出A个杯子，一开始都朝上，每次可以翻B个杯子，问最少需要翻转多少次可以让所有杯子都朝下。分类讨论：首先对于A%B==0一类情况，直接输出。对于A>=3B，让A减到[2B,3B)区间内，翻转次数累加上A/B-2。当A>=2B时，分奇偶讨论：A为奇数B为偶数显然无解；AB同奇偶时最多需要3次，A偶数B奇数最多需要4次。当A<2B时，分奇偶讨论：AB同奇偶时最多需要3次，A奇数B偶数无解，A偶数B奇数时，有F(A,B)=F(A,A-B)成立，可以转换成上面的情况求解即可。具体证明画画图就知道了，将两个B分别放到对称的位置上，想办法调整使得每阅读全文

posted @ 2011-08-01 16:25 kuangbin 阅读(352) 评论(0) 推荐(0) 编辑

ACM HDU 1098Ignatius's puzzle

摘要： Ignatius's puzzleTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2716Accepted Submission(s): 1788Problem DescriptionIgnatius is poor at math,he falls across a puzzle problem,so he has no choice but to appeal to Eddy. this problem describes that 阅读全文

posted @ 2011-08-01 10:49 kuangbin 阅读(643) 评论(0) 推荐(0) 编辑