[hdu 4945]14多校第八场A 2048 状压DP+数论

2048

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 489    Accepted Submission(s): 112

Problem Description

Teacher Mai is addicted to game 2048. But finally he finds it's too hard to get 2048. So he wants to change the rule:

You are given some numbers. Every time you can choose two numbers of the same value from them and merge these two numbers into their sum. And these two numbers disappear meanwhile.
  
If we can get 2048 from a set of numbers with this operation, Teacher Mai think this multiset is good.

You have n numbers, A ₁,...,A _n. Teacher Mai ask you how many subsequences of A are good.

The number can be very large, just output the number modulo 998244353.

 

Input

There are multiple test cases, terminated by a line "0".

For each test case, the first line contains an integer n (1<=n<=10^5), the next line contains n integers a _i (0<=a _i<=2048).

 

Output

For each test case, output one line "Case #k: ans", where k is the case number counting from 1, ans is the number module 998244353.

 

Sample Input

   
   
   
   

    
    
    
    4
1024 512 256 256
4
1024 1024 1024 1024
5
1024 512 512 512 1
0
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case #1: 1
Case #2: 11
Case #3: 8


    
    
    
    
 
     
 
      Hint 
     
In the first case, we should choose all the numbers. In the second case, all the subsequences which contain more than one number are good. 
    
    
    
    
     
   
   
   
   

 

题目大意

给定一个数的集合，问其中有多少子集满足：

两个相同的数字可以加在一起，重复若干次这样的操（可为0次）作得到2048。结果对998244353取模

解题思路

去除不为2的整数次幂的所有数字（设有fr个），将最后结果乘上2^fr（是否取这些数字不影响2048的操作）

枚举

dp[i][j]表示取到第i个2^i的数,其最大的和在j*2^i至(j+1)*2^i-1的方案数

答案就是可的总方案减去dp[12][0]  (即结果落在0~2047的方案个数)

对于每个dp[i][j]

我们有枚举下一步状态时dp[i+1][k]增加的方案数为dp[i][j]*C(cnt[i],k-j/2) 

其中，cnt[i]为2^i出现的次数，C为组合数，998244353是质数所以组合数采用逆元处理。

PS：

原题标答自带的输入优化跑了500ms……

读入优化之后……

G++反复提交通不过

C++1A……1390ms……

code:

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <ctime>
#include <cctype>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <queue>
#include <list>
#include <vector>
#include <map>
#include <set>

#define sqr(x) ((x)*(x))
#define LL long long 
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define eps 1e-10
#define mod 998244353ll
using namespace std;
inline int ReadInt()
{
    int flag=0;
    char ch=getchar();
    int data=0;
    while (ch<'0'||ch>'9')
    {
        if (ch=='-') flag=1;
        ch=getchar();
    }
    do
    {
        data=data*10+ch-'0';
        ch=getchar();
    }while (ch>='0'&&ch<='9');
    return data;
}
void egcd(LL a,LL b,LL &x,LL &y)
{
	if (b==0)
	{
		x=1,y=0;
		return ;
	}
	egcd(b,a%b,x,y);
	LL t=x;
	x=y;y=t-a/b*x;
}
int a[100005];
LL inv[100005];
LL cnt[15],fr;
int base[3000];
LL dp[20][2050];
int lim[20];
LL mypow(LL x,LL y)
{
	LL res=1;
	LL mul=x;
	while (y)
	{
		if (y&1)
		res=res*x%mod;
		x=x*x%mod;
		y>>=1;
	}
	return res;
}
int main()
{
	lim[0]=2048;
	for (int i=1;i<=12;i++)
	lim[i]=2048>>(i-1);
	for (LL i=1;i<=100000;i++)
	{
		LL x,y;
		egcd(i,mod,x,y);
		x=(x+mod)%mod;
		inv[i]=x;
	}
	int n,ca=0;
	while (~scanf("%d",&n),n)
	{
		fr=n;
		memset(base,0,sizeof base);
		for (int i=1;i<=n;i++){
		scanf("%d",&a[i]);
		base[a[i]]++;
		}
		memset(cnt,0,sizeof cnt);
		for (int i=1;i<=12;i++)
		{
			cnt[i]=base[1<<(i-1)];
			fr-=cnt[i];
		}
		memset(dp,0,sizeof dp);
		dp[0][0]=1;
		for (int i=1;i<=12;i++)
		{
			for (int j=0;j<=lim[i-1];j++)
			if (dp[i-1][j])
			{
				LL cal=1;
				for (int k=j/2;k<lim[i];k++)
				{
					int pos=j/2;
					if ((k-pos)>cnt[i]) break;
					dp[i][k]+=dp[i-1][j]*cal%mod;
					dp[i][k]%=mod;
					cal=(cal*(LL)(cnt[i]-((k-pos))))%mod;
					cal=(cal*inv[(k-pos)+1])%mod;
				}
			}
		}

		LL ans;
		ans=(mypow(2ll,(LL)n-fr)-dp[12][0]+mod)%mod;
		ans=ans*mypow(2ll,fr)%mod;
		printf("Case #%d: %I64d\n",++ca,ans);
	}
	return 0;
}