题目大意：给你一个序列让你求出有多少种组合可以得到2048.结果要对998244353取余。
解题思路：求出不能满足条件的方案数，然后用总的减去不满足的然后乘上其他无关的组合方式，比如3，5这些数字是在构成2048的过程中无用的，所以乘上这些组合出来的情况。
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。
所以有dp[i][j] += ((dp[i-1][k]+dp[i-1][k+1])*use[i][j-k/2])%mod。表示在i位置时最大和为j的方案数，有一部分是之前已经得到的，一部分是第i次新添加的。
这道题目很卡时间，所以要组合数的取模要用到逆元然后加快速幂取模就可以了啊。
PS：感觉挺好的题目，以后要多做做。
2048
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1052 Accepted Submission(s): 276
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, A1,...,An. 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 ai (0<=ai<=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.
Author
xudyh
Source
2014 Multi-University Training Contest 8
#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-10
///#define M 1000100
///#define LL __int64
#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)
using namespace std;
const int maxn = 100010;
#define mod 998244353
int num[21];
LL dp[15][2100];
LL use[15][2100];
LL f[maxn];
int n;
int add(int x)
{
if(x == 0) return 20;
int ans = 0;
while(x!=1)
{
if(x&1) return 20;
x >>= 1;
ans++;
}
return ans;
}
LL q_mod(LL a,LL b,LL n)
{
LL ret=1;
LL tmp=a%n;
while(b)
{
//基数存在
if(b&0x1) ret=ret*tmp%n;
tmp=tmp*tmp%n;
b>>=1;
}
return ret;
}
LL Del(int n, int m)
{
if(m > n) return 0LL;
LL ans = f[n];
ans = (ans*q_mod((f[m]*f[n-m])%mod, mod-2, mod))%mod;
return ans;
}
void select()
{
for(int i = 1; i <= 11; i++)
for(int j = 0; j < 2048; j++) use[i][j] = Del(num[i-1], j);
}
void change()
{
select();
for(int i = 0; i < 2048; i++) dp[1][i] = use[1][i];
for(int i = 2; i <= 11; i++)
{
int cnt = (1<<(12-i));
for(int j = 0; j < cnt; j++)
{
for(int k = 0; k <= 2*j+1; k += 2)
{
if(!use[i][j-k/2] || !(dp[i-1][k]+dp[i-1][k+1])) continue;
dp[i][j] += ((dp[i-1][k]+dp[i-1][k+1])*use[i][j-k/2])%mod;
}
dp[i][j] %= mod;
}
}
LL ans = 0LL;
ans = q_mod(2, num[20], mod);
LL ps = (dp[11][0]+dp[11][1])%mod;
ps = q_mod(2, n-num[20], mod)-ps;
ps = (ps+mod)%mod;
ans = (ans*ps)%mod;
printf("%I64d\n",ans);
}
void init()
{
f[0] = 1LL;
for(LL i = 1; i < maxn; i++) f[i] = (f[i-1]*i)%mod;
}
int main()
{
int Case = 1;
init();
while(~scanf("%d",&n) && n)
{
memset(num, 0, sizeof(num));
memset(dp, 0, sizeof(dp));
int x;
for(int i = 0; i < n; i++)
{
scanf("%d",&x);
num[add(x)]++;
}
printf("Case #%d: ",Case++);
change();
}
return 0;
}