hdu5816 Hearthstone(状压DP)
来源:互联网 发布:淘宝营业执照代办 编辑:程序博客网 时间:2024/05/17 03:50
思路:令dp[s]为当前手牌为s状态时候的方案数,那么假设手上有x张A,y张B,那它的下一个状态就是2*x-(x+y)+1张牌,然后从小到大枚举S更新状态,当前方案贡献的排列数就是dp[s]*(n+m-x-y)! 不过这样会T成狗,我预处理了一下阶乘表以及预处理了0到1<<20的所有数的1的出现次数,然后就可以了
#include<bits/stdc++.h>using namespace std;#define LL long longint n,m,h;int countt[(1<<20)+5];LL f[22]={1};LL gcd(LL a,LL b){return a==0?b:gcd(b%a,a);}void init(){for(int i = 0;i<=(1<<20);i++)for(int j = 0;j<20;j++)if(i&(1<<j))countt[i]++;for(int i = 1;i<=20;i++)f[i]=f[i-1]*i;}int a[22];LL dp[(1<<20)+5];int cal(int s){int ans = 0;for(int i = 0;i<m;i++)if(s&(1<<i))ans+=a[i];return ans;}int coun(int s){int ans = 0;for(int i = 0;i<m;i++)if(s&(1<<i))ans++;return 2*(countt[s]-ans)+1-countt[s];}int main(){init();int T;scanf("%d",&T);while(T--){memset(dp,0,sizeof(dp));scanf("%d%d%d",&h,&n,&m);int tot = n+m;int all = (1<<tot)-1; for(int i = 0;i<m;i++)scanf("%d",&a[i]);LL ans = 0;if(cal(all)>=h){dp[0]=1; for(int i = 0;i<(1<<tot);i++){if(dp[i]){ if(i==all || coun(i)==0){if(cal(i)>=h)ans+=dp[i]*f[tot-countt[i]];}else{for(int j = 0;j<tot;j++)if(!(i&(1<<j)))dp[i|(1<<j)]+=dp[i];}}}}LL gg = gcd(ans,f[tot]);printf("%lld/%lld\n",ans/gg,f[tot]/gg);}}
Problem Description
Hearthstone is an online collectible card game from Blizzard Entertainment. Strategies and luck are the most important factors in this game. When you suffer a desperate situation and your only hope depends on the top of the card deck, and you draw the only card to solve this dilemma. We call this "Shen Chou Gou" in Chinese.
Now you are asked to calculate the probability to become a "Shen Chou Gou" to kill your enemy in this turn. To simplify this problem, we assume that there are only two kinds of cards, and you don't need to consider the cost of the cards.
-A-Card: If the card deck contains less than two cards, draw all the cards from the card deck; otherwise, draw two cards from the top of the card deck.
-B-Card: Deal X damage to your enemy.
Note that different B-Cards may have different X values.
At the beginning, you have no cards in your hands. Your enemy has P Hit Points (HP). The card deck has N A-Cards and M B-Cards. The card deck has been shuffled randomly. At the beginning of your turn, you draw a card from the top of the card deck. You can use all the cards in your hands until you run out of it. Your task is to calculate the probability that you can win in this turn, i.e., can deal at least P damage to your enemy.
Now you are asked to calculate the probability to become a "Shen Chou Gou" to kill your enemy in this turn. To simplify this problem, we assume that there are only two kinds of cards, and you don't need to consider the cost of the cards.
-A-Card: If the card deck contains less than two cards, draw all the cards from the card deck; otherwise, draw two cards from the top of the card deck.
-B-Card: Deal X damage to your enemy.
Note that different B-Cards may have different X values.
At the beginning, you have no cards in your hands. Your enemy has P Hit Points (HP). The card deck has N A-Cards and M B-Cards. The card deck has been shuffled randomly. At the beginning of your turn, you draw a card from the top of the card deck. You can use all the cards in your hands until you run out of it. Your task is to calculate the probability that you can win in this turn, i.e., can deal at least P damage to your enemy.
Input
The first line is the number of test cases T (T<=10).
Then come three positive integers P (P<=1000), N and M (N+M<=20), representing the enemy’s HP, the number of A-Cards and the number of B-Cards in the card deck, respectively. Next line come M integers representing X (0<X<=1000) values for the B-Cards.
Then come three positive integers P (P<=1000), N and M (N+M<=20), representing the enemy’s HP, the number of A-Cards and the number of B-Cards in the card deck, respectively. Next line come M integers representing X (0<X<=1000) values for the B-Cards.
Output
For each test case, output the probability as a reduced fraction (i.e., the greatest common divisor of the numerator and denominator is 1). If the answer is zero (one), you should output 0/1 (1/1) instead.
Sample Input
23 1 21 23 5 101 1 1 1 1 1 1 1 1 1
Sample Output
1/346/273
Author
SYSU
Source
2016 Multi-University Training Contest 7
0 0
- hdu5816 Hearthstone(状压DP)
- hdu5816 Hearthstone 状压dp
- hdu5816 Hearthstone 【状压dp】
- 【HDU5816】Hearthstone(状压DP)
- hdu5816 Hearthstone 状态压缩dp
- hdu5816 多校7 Hearthstone【组合计数+dp】
- HDU5816 Hearthstone
- hdu5816 状压dp
- hdu 5816 Hearthstone (状压dp)
- [HDU 5816] Hearthstone (概率DP+状压)
- HDU 5816 Hearthstone (状压dp)
- HDU 5816 Hearthstone(概率DP+状压)
- HDU 5816 Hearthstone (状压dp)
- HDU 5816 Hearthstone (状压dp+概率)
- HDU 5816 Hearthstone(状压DP)
- 2016多校训练Contest7: 1008 Hearthstone hdu5816
- HDU5816(状压DP,位运算的一些技巧)
- CSU1646: HearthStone(DP)
- Vsftpd使用FtpClient上传文件踩过的坑
- 我自己本机配置了一个maven环境。每次新开一个项目都要重新选择一个maven的配置,默认的都是这个。如图:
- 机器学习-线性回归-正规方程
- HTML5(李炎恢)学习笔记二 ------------- HTML5的结构
- machine-learning-ex2_1
- hdu5816 Hearthstone(状压DP)
- 磁条卡驱动
- absolute绝对定位的非绝对定位用法 张鑫旭博客
- SSRF Tips
- 初心
- Moscow Subregional 2010 Problem K. KMC Attacks 交互题、队列优化、枚举
- 从零开始学_JavaScript_系列(27)——myblog的优化【1】样式表分离、localStorage
- Qt Creator插件制作小插曲:有关QT_NO_CAST_FROM_ASCII的注意事项
- 模拟指针学习笔记(1)-实现模拟指针系统