hdu4945(dp+组合数学）

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题意：给一串数字，问有多少个序列能够组成2048

思路分析：DP+组合数学，如果就一个DP就还好做，加上组合数学，那我就晕了。思路是参考大神的。dp[i][j]表示2^i选了j个有多少个序列，状态转移dp[i][j+k/2] += C(n,j)*dp[i-1][k]。还有组合数是怎么用的很关键，看代码吧。

这题非常卡时间，有两点注意下：1、这题的输入数据量很大，要进行输入优化；2、取模运算非常耗时，尽量少用取模运算，我用add来减少取模运算

代码如下：

#include<iostream>#include<algorithm>#include<cstring>#include<string>#include<stack>#include<queue>#include<set>#include<map>#include<stdio.h>#include<stdlib.h>#include<math.h>#define LL __int64#define MOD 998244353#define N 100005#define M 2050#define inf 0x7ffffff#define eps 1e-9#define pi acos(-1.0)using namespace std;int a[20],vis[M],cnt_c[20];LL p[N],pw[N],pm[N];LL dp[20][M];LL pow_mul(LL a,int n){    LL b = 1;    while(n)    {        if(n&1) b = b*a%MOD;        n >>= 1;        a = a*a%MOD;    }    return b;}void init(){    memset(vis,-1,sizeof(vis));    int i;    for(i = 0; i < 12; i++)    {        vis[1<<i] = i;        cnt_c[i] = 2048/(1<<i);    }    p[0] = 1;    for(i = 1; i < N; i++)        p[i] = p[i-1]*2%MOD;    pm[0] = pw[0] = 1;    for(i = 1; i < N; i++)    {        pm[i] = pm[i-1]*i%MOD;        pw[i] = pow_mul(pm[i],MOD-2);    }}LL C(int n,int m){    return pm[n]*pw[m]%MOD*pw[n-m]%MOD;}void add(LL &a,LL b){    a += b;    if(a >= MOD) a-= MOD;}void solve(){    memset(dp,0,sizeof(dp));    int i,j,k;    int num = min(a[0],cnt_c[0]);    LL sum = 0,val;    for(i = 0; i <= num; i++){        val = C(a[0],i);        dp[0][i] = val;        add(sum,val);    }    if(cnt_c[0] < a[0])        dp[0][cnt_c[0]] = (dp[0][cnt_c[0]] + p[a[0]] - sum + MOD)%MOD;    for(i = 1; i < 12; i++)    {        num = min(a[i],cnt_c[i]);        sum = 0;        for(j = 0; j <= num; j++)        {            val = C(a[i],j);            add(sum,val);            for(k = 0; k <= cnt_c[i-1]; k++)                if(dp[i-1][k])                {                    int temp = min(j+k/2,cnt_c[i]);                    add(dp[i][temp],val*dp[i-1][k]%MOD);                }        }        if(a[i] > cnt_c[i]){            val = (p[a[i]]-sum + MOD)%MOD;            for(j = 0; j <= cnt_c[i-1]; j++)                if(dp[i-1][j])                        add(dp[i][cnt_c[i]],dp[i-1][j]*val%MOD);        }    }}void read(int &x){    char ch;    while((ch=getchar())<'0'||ch>'9');    x=ch-'0';    while((ch=getchar())>='0'&&ch<='9')        x=x*10+ch-'0';}int main(){//freopen("input.txt","r",stdin);//freopen("output.txt","w",stdout);    int n,cnt,cas = 1;    init();    while(scanf("%d",&n) && n)    {        int i;        cnt = 0;        memset(a,0,sizeof(a));        for(i = 0; i < n; i++)        {            int x;            read(x);            if(vis[x] == -1) cnt++;            else a[vis[x]]++;        }        printf("Case #%d: ",cas++);        solve();        printf("%I64d\n",dp[11][1]*p[cnt]%MOD);    }    return 0;}

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