ICPCCamp 2016 Day 3 - Grand Prix of China(Random Arithmetic-dp)

题意：已知n个数，每次等概率随机选取2个数，等概率相加或相乘，并用这个数代替所取的2个数，直到只剩一个数为止，问该数期望。

2个数的情况 x,y
ans=x+y+xy2

显然对于a1,a2,…,an而言，答案为∑Lf(n,i)al1al2…ali 其中L为A的任意子集，其中的元素为l1,l2,…,ln,f(n,i)为只与|A|,|L|有关函数。

那么现在可以递归求f(n,i)
假设w1,w2是L中2元素,x,y是不在L中的2元素，则有4种情况。

• xy
• w1w2
• w1+w2
• x+w

分类讨论

#include <iostream>#include <cmath>#include <algorithm>#include <cstdio>#include <cstring>#include <string>#include <vector>#include <map>#include <functional>#include <cstdlib>#include <queue>#include <stack>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=Pre[x];p;p=Next[p])#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (1000000007)#define pb push_back#define mp make_pair #define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())typedef long long ll;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} int n;const int MAXN = ( 2000 + 10 );ll f[MAXN][MAXN]={0};ll pow2(ll a,ll b){    if (b==1) return a;    if (b==0) return 1;    ll t= pow2(a,b/2);    t =t *t%F;    if (b&1) t=t*a%F;    return t;}ll inv(ll x) {return pow2(x,F-2);}void prework() {    Rep(i,n+1) f[i][0]=1;    For(i,n) For(x,i) {   // (x,y) w        if (i==1&&x==1) {            f[1][1]=1;continue;        }        ll p1=mul(mul(mul(mul(x,i-x) , inv(i) ),inv(i-1)) ,2); //跨立        ll p2=mul(mul(mul(x,x-1) , inv(i) ),inv(i-1)) ; // 2 个里面的        ll p3=mul(mul(mul(i-x,i-x-1) , inv(i) ),inv(i-1) ) ; // 2个外面的        p1=mul(p1,inv(2));        p2=mul(p2,inv(2));        p3=mul(p3,inv(2));        if (x>=2) upd(f[i][x],mul(p2,f[i-1][x-1])); //(xy)         if (i-x>=2) upd(f[i][x],mul(p3,f[i-1][x])); //w1w2        if (i-x>=2) upd(f[i][x],mul(p3,f[i-1][x])); // w1+w2        if (i-x>=1&&x>=1) upd(f[i][x],mul(p1,f[i-1][x])); // x+w    }}ll S[MAXN][MAXN]={0};ll a[MAXN];int main() {    n=read();    prework();    For(i,n) a[i]=read();    S[0][0]=1; For(i,n) S[i][0]=1;    ll ans=0;    For(i,n) For(j,i) {        if (j>=1) upd(S[i][j],mul(S[i-1][j-1],a[i] ) );        if (i>j) upd(S[i][j], S[i-1][j] );    }    For(i,n) {        upd(ans, mul(S[n][i] , f[n][i] ));    }    cout<<ans<<endl;    return 0;}