ZOJ 4257 MostPowerful（状压DP，简单）

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CF895C Square Subsets (组合数+状压DP+简单数论)

题目大意:给你一个序列,你可以在序列中任选一个子序列,求子序列每一项的积是一个平方数的方案数. 1<=a[i]<=70 因为任何一个大于2的数都可以表示成几个质数的幂的乘积所以我们预处理70以内的质数,把它作为二进制状压的状态,每个在序列中出现数Hash一下,组合数推一下所以把奇次幂的状态表示为1,偶次幂的状态就是0,比如6就是11,42就是1011 而平方数的每个质因子的指数都是偶数,所以最终结果的状态就是0000000... 转移的过程,两个数的乘积,就是这两个数的质因子二进制的状态的…

ZOJ 3777-Problem Arrangement(状压DP)

B - Problem Arrangement Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu Submit Status Practice ZOJ 3777 Description The 11th Zhejiang Provincial Collegiate Programming Contest is coming! As a problem setter, Edward is going…

2019.02.09 bzoj2560: 串珠子（状压dp+简单容斥）

传送门题意简述:nnn个点的带边权无向图,定义一个图的权值是所有边的积,问所有nnn个点都连通的子图的权值之和. 思路: fif_ifi表示保证集合iii中所有点都连通其余点随意的方案数. gig_igi表示只考虑集合iii中所有点的状态的子图的权值和. 我们先预处理出ggg数组,然后考虑递推fff数组. 显然fif_ifi是等于gig_igi扣掉一些东西的,扣掉的应该就是不连通的情况. 于是我们枚举编号最小的点所在的连通块来扣掉非法情况. 时间复杂度O(n2n+3n)O(n2^n+3…

ZOJ 3777 - Problem Arrangement - [状压DP][第11届浙江省赛B题]

题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3777 Time Limit: 2 Seconds Memory Limit: 65536 KB The 11th Zhejiang Provincial Collegiate Programming Contest is coming! As a problem setter, Edward is going to arrange the order…

Problem Arrangement ZOJ - 3777(状压dp + 期望)

ZOJ - 3777 就是一个入门状压dp期望 dp[i][j] 当前状态为i,分数为j时的情况数然后看代码有注释 #include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <cctype> #include <set> #include <vector> #inclu…

ZOJ - 3777(状压dp)

The 11th Zhejiang Provincial Collegiate Programming Contest is coming! As a problem setter, Edward is going to arrange the order of the problems. As we know, the arrangement will have a great effect on the result of the contest. For example, it will…

ZOJ 3306 状压dp

转自:http://blog.csdn.net/a497406594/article/details/38442893 Kill the Monsters Time Limit: 7 Seconds Memory Limit: 32768 KB In order to celebrate the 8th anniversary of ZOJ, watashi introduces a strange game to other ZJU ACM team members. The board of…