题目链接:HDU 5119 Problem Description Matt has N friends. They are playing a game together. Each of Matt's friends has a magic number. In the game, Matt selects some (could be zero) of his friends. If the xor (exclusive-or) sum of the selected friends'ma…

Matt has N friends. They are playing a game together.Each of Matt's friends has a magic number. In the game, Matt selects some (could be zero) of his friends. If the xor (exclusive-or) sum of the selected friends'magic numbers is no less than M , Mat…

虽然是一道还是算简单的DP,甚至不用滚动数组也能AC,数据量不算很大. 对于N个数,每个数只存在两个状态,取 和 不取. 容易得出状态转移方程: dp[i][j] = dp[i - 1][j ^ a[i]] + dp[i - 1][j]; dp[i][j] 的意思是,对于数列 中前 i 个数字,使得 XOR 和恰好为 j 的方案数 状态转移方程中的 dp[i - 1][j] 即表示当前这个数字不取, dp[i - 1][j ^ a[i]] 表示当前这个数字要取. 这道题还是要好好理解阿! sou…

http://acm.hdu.edu.cn/showproblem.php?pid=5119 题意:给出n个数和一个上限m,求从这n个数里取任意个数做异或运算,最后的结果不小于m有多少种取法. 思路:dp[i][j]表示在前i个数中取数做异或运算最后结果为j的方法数,那么dp[i][j] = dp[i-1][j] + dp[i-1][j^a[i]],分别对于取与不取这两种状态. 因为要推导出第i的状态只有i-1有关,所以这里可以用滚动数组. #include<cstdio> #include&…

Happy Matt Friends Time Limit: 6000/6000 MS (Java/Others)    Memory Limit: 510000/510000 K (Java/Others)Total Submission(s): 5188    Accepted Submission(s): 1985 Problem Description Matt has N friends. They are playing a game together. Each of Matt’s…

Happy Matt Friends Time Limit: 6000/6000 MS (Java/Others) Memory Limit: 510000/510000 K (Java/Others) Total Submission(s): 3215 Accepted Submission(s): 1261 Problem Description Matt has N friends. They are playing a game together. Each of Matt's frie…

Description Yesterday your dear cousin Coach Pang gave you a new 100MB hard disk drive (HDD) as a gift because you will get married next year. But you turned on your computer and the operating system (OS) told you the HDD is about 95MB. The 5MB of sp…

Problem Description 电子科大本部食堂的饭卡有一种很诡异的设计,即在购买之前判断余额.如果购买一个商品之前,卡上的剩余金额大于或等于5元,就一定可以购买成功(即使购买后卡上余额为负),否则无法购买(即使金额足够).所以大家都希望尽量使卡上的余额最少.某天,食堂中有n种菜出售,每种菜可购买一次.已知每种菜的价格以及卡上的余额,问最少可使卡上的余额为多少.   Input 多组数据.对于每组数据:第一行为正整数n,表示菜的数量.n<=1000.第二行包括n个正整数,表示每种菜的价格…

题意:给定n个数,从中分别取出0个,1个,2个...n个,并把他们异或起来,求大于m个总的取法. 思路:dp,背包思想,考虑第i个数,取或者不取,dp[i][j]表示在第i个数时,异或值为j的所有取法.dp[i][j] = dp[i - 1][j] + dp[i - 1][j ^ a[i]]); 其中dp[i - 1][j]表示不取第i个数,dp[i - 1][j ^ a[i]]表示取第i个数,由于40比较大,所以用滚动数组优化,后一个状态需要前一个来推导,而和前一个之前的所有的没有关系,所以之…

题目链接 题意:n个数,你可以从中选一些数,也可以不选,选出来的元素的异或和大于m时,则称满足情况.问满足情况的方案数为多少. 分析:本来以为是用什么特殊的数据结构来操作,没想到是dp,还好队友很强.定义dp[i][j]为在前i个数里选一些数的异或和为j的方案数,边计算边统计, #include <iostream> #include <cstdio> #include <cstring> typedef long long ll; <<); using n…