/* 2^x mod n = 1 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 11800    Accepted Submission(s): 3673 Problem Description Give a number n, find the minimum x(x>0) that satisfies 2^x mod n = 1.…

循环多少次? Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 2839    Accepted Submission(s): 1028 Problem Description   我们知道,在编程中,我们时常须要考虑到时间复杂度.特别是对于循环的部分. 比如, 假设代码中出现 for(i=1;i<=n;i++) OP ; 那么做了n次O…

/* Lowest Bit Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 7775    Accepted Submission(s): 5714 Problem Description Given an positive integer A (1 <= A <= 100), output the lowest bit of A.…

/* 最小公倍数 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 32933    Accepted Submission(s): 18398 Problem Description 给定两个正整数,计算这两个数的最小公倍数. Input 输入包括多组測试数据,每组仅仅有一行,包括两个不大于1000的正整数.   Output 对于每一…

/* 放大的X Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 8866    Accepted Submission(s): 2678 Problem Description 请你编程画一个放大的'X'. 如3*3的'X'应例如以下所看到的: X X X X X 5*5的'X'例如以下所看到的: X      X X X X X X…

/* 大菲波数 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 11168    Accepted Submission(s): 3782 Problem Description Fibonacci数列.定义例如以下: f(1)=f(2)=1 f(n)=f(n-1)+f(n-2) n>=3. 计算第n项Fibonacci数值. Inpu…

/* hdu 1999 不可摸数 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 8274    Accepted Submission(s): 2122 Problem Description s(n)是正整数n的真因子之和.即小于n且整除n的因子和.比如s(12)=1+2+3+4+6=16.假设不论什么 数m.s(m)都不等于n,则…

HDOJ 题目分类 //分类不是绝对的 //"*" 表示好题,需要多次回味 //"?"表示结论是正确的,但还停留在模块阶 段,需要理解,证明. //简单题看到就可以敲的 1000:    入门用: 1001:    用高斯求和公式要防溢出 1004:1012: 1013:    对9取余好了 1017:1021: 1027:    用STL中的next_permutation() 1029:1032:1037:1039:1040:1056:1064:1065: 10…

原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=6655 简单博弈问题,A,B手里各有n,m张牌,牌有颜色,两人轮流出牌(A先出),一个人只能打出对放未打出过的颜色的牌(可以打出自己打出过的颜色的牌),当一方不能再打出牌时,对方获胜. 博弈策略:优先选择双方都有该颜色的牌(对方没有的颜色和自己没有的颜色放在最后,不影响),再优先选择双方牌数相加最大的颜色的牌. 存储方法:读入之后离散化,统计数量,放入结构体数组,最后排序 #include <iost…

BestCoder Sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 573    Accepted Submission(s): 201 Problem Description Mr Potato is a coder.Mr Potato is the BestCoder. One night, an amazing s…