杰拉斯的博客

标签:算法

[ACM_HDU_1410]PK武林盟主

杰拉斯 杰拉斯 | 时间:2012-04-20, Fri | 21,579 views
编程算法 

PK武林盟主

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 621 Accepted Submission(s): 133

Description

枫之羽认为自己很强,想当武林盟主,于是找现任武林盟主氢氧化铜挑战。氢氧化铜欣然接受了挑战,两人约好于下个月的月圆之夜在HDU校园内的三根柱子上进行决战。这场PK赛肯定能吸引武林中所有人前来观战,所以他们找了有商业运作潜力的经济人你,让你来组织这场百年一见的世纪之战,假设两人都有一定的血HP1、HP2.HP1是枫之羽的,HP2是氢氧化铜的。他们也有一定攻击力AP1、AP2,AP1是枫之羽的,AP2是氢氧化铜的。当进行攻击时,对方的HP减少自己的攻击力,比如HP1=2 HP2=1 AP1=1 AP2=1,当氢氧化铜攻击枫之羽时,枫之羽的HP=2(原先的HP1)-1(氢氧化铜的AP2)=1。现在两个人对决很多回合,每回合不是枫之羽攻击氢氧化铜,就是氢氧化铜攻击枫之羽。求枫之羽能赢氢氧化铜成为下任武林盟主的的胜率。

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[ACM_NYOJ_270]Product of Digits

杰拉斯 杰拉斯 | 时间:2012-04-20, Fri | 18,224 views
编程算法 

Product of Digits

Time Limit: 1.0 second
Memory Limit: 16 MB

Description

Your task is to find the minimal positive integer number Q so that the product of digits of Q is exactly equal to N.

Input

The input contains the single integer number N (0 ≤ N ≤ 109).

Output

Your program should print to the output the only number Q. If such a number does not exist print −1.

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[ACM_HDU_1009]FatMouse' Trade(贪心入门)

杰拉斯 杰拉斯 | 时间:2012-04-18, Wed | 22,176 views
编程算法 

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22404 Accepted Submission(s): 6972

Description

FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean.
The warehouse has N rooms. The i-th room contains J[i] pounds of JavaBeans and requires F[i] pounds of cat food. FatMouse does not have to trade for all the JavaBeans in the room, instead, he may get J[i]* a% pounds of JavaBeans if he pays F[i]* a% pounds of cat food. Here a is a real number. Now he is assigning this homework to you: tell him the maximum amount of JavaBeans he can obtain.

Input

The input consists of multiple test cases. Each test case begins with a line containing two non-negative integers M and N. Then N lines follow, each contains two non-negative integers J[i] and F[i] respectively. The last test case is followed by two -1's. All integers are not greater than 1000.

Output

For each test case, print in a single line a real number accurate up to 3 decimal places, which is the maximum amount of JavaBeans that FatMouse can obtain.

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[ACM_NYOJ_15]括号匹配(二)

杰拉斯 杰拉斯 | 时间:2012-04-18, Wed | 18,486 views
编程算法 

括号匹配(二)

时间限制:1000 ms | 内存限制:65535 KB
难度:6

描述

给你一个字符串,里面只包含"(",")","[","]"四种符号,请问你需要至少添加多少个括号才能使这些括号匹配起来。
如:
[]是匹配的
([])[]是匹配的
((]是不匹配的
([)]是不匹配的

输入

第一行输入一个正整数N,表示测试数据组数(N<=10)
每组测试数据都只有一行,是一个字符串S,S中只包含以上所说的四种字符,S的长度不超过100

输出

对于每组测试数据都输出一个正整数,表示最少需要添加的括号的数量。每组测试输出占一行

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[ACM_NYOJ_2]括号匹配(一)

杰拉斯 杰拉斯 | 时间:2012-04-18, Wed | 17,095 views
编程算法 

括号配对问题

时间限制:3000 ms | 内存限制:65535 KB
难度:3

描述

现在,有一行括号序列,请你检查这行括号是否配对。

输入

第一行输入一个数N(0<N<=100),表示有N组测试数据。后面的N行输入多组输入数据,每组输入数据都是一个字符串S(S的长度小于10000,且S不是空串),测试数据组数少于5组。数据保证S中只含有"[","]","(",")"四种字符

输出

每组输入数据的输出占一行,如果该字符串中所含的括号是配对的,则输出Yes,如果不配对则输出No。

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