CF453B CF454D Codeforces Round #259 (Div. 2) D Codeforces Round #259 (Div. 1) B D. Little Pony and Harmony Chest time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output Princess Twilight went to C…

Little Pony and Harmony Chest 经典状态压缩dp #include <cstdio> #include <cstring> #include <cstdlib> #include <iostream> #define min(x,y) (x>y?y:x) using namespace std; ],all,n,a[],b[][<<],pre[][<<],s1,s0,f[][<<],ans…

D. Little Pony and Harmony Chest   Princess Twilight went to Celestia and Luna's old castle to research the chest from the Elements of Harmony. A sequence of positive integers bi is harmony if and only if for every two elements of the sequence their…

2101: [Usaco2010 Dec]Treasure Chest 藏宝箱 Time Limit: 10 Sec  Memory Limit: 64 MBSubmit: 327  Solved: 147[Submit][Status] Description Bessie and Bonnie have found a treasure chest full of marvelous gold coins! Being cows, though, they can't just walk i…

The square chest Sophia pressed the button in front of her, slamming her fist against it. The door rumbled and appeared to do nothing. “Oh man, that’s not very interesting at all.” Sofia complained. Suddenly, the the door hissed and a jet of ice cold…

dp( l , r ) = sum( l , r ) - min( dp( l + 1 , r ) , dp( l , r - 1 ) ) 被卡空间....我们可以发现 l > r 是无意义的 , 所以可以省下一半的空间 -------------------------------------------------------------------------------- #include<cstdio> #include<cstring> #include<a…

G - Zombie’s Treasure Chest Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Description   Some brave warriors come to a lost village. They are very lucky and find a lot of treasures and a big treasure chest,…

 Zombie's Treasure Chest 本题题意:有一个给定容量的大箱子,此箱子只能装蓝宝石和绿宝石,假设蓝绿宝石的数量无限,给定蓝绿宝石的大小和价值,要求是获得最大的价值 题解:本题看似是dp中的背包问题,但是由于数据量太大,用dp肯定会超时,所以只能寻找另外一种思路,可以用贪心加暴力,先求出两种宝石大小的最小公倍数com,然后将N/com-com,与N%comkanchengs看成是两个部分(想想应该明白).将前一个部分,放入单位价值量最高的那个,对于后面那个部分直接将S1的数量从…

[CF453B]Little Pony and Harmony Chest 题目大意: 给你一个长度为\(n(n\le100)\)的正整数序列\(A(A_i\le30)\),求一个正整数序列\(B\),使得\(\sum_{i=1}^n |A_i-B_i|\)最小,且\(B\)中所有互质. 思路: 由于\(B_i\ge59\)时用\(1\)代替时不会更差,因此我们只需要考虑\(B_i\le58\)的情况.由于质因数只有\(16\)个,因此可以状压DP.另外记录转移即可. 源代码: #include…

454D - Little Pony and Harmony Chest 思路: 状压dp,由于1的时候肯定满足题意,而ai最大是30,所以只要大于等于59都可以用1替换,所以答案在1到59之间 然后筛出1到58之间的质数,只有16个,把1到58的数的状态由这16个质数表示,如果整除这个质数则二进制中这一位为1,否则则为0 状态:dp[i][j]表示到第i个数为止选取的数的状态为j的最小差和 初始状态:dp[0][0]=0 状态转移: dp[i+1][j|sta[k]]=min(dp[i+1][…