Codeforces Good Bye 2017 D. New Year and Arbitrary Arrangement

首先考 慮dp [i] [j] 表示 i 個a j個子序列ab 時候執行算法的期望

那麼轉移就是dp[i][j] = ( pa * dp [i+1][j] + pb * dp[i][i+j] ) / (pa + pb)

因爲長度可以任意長所以首先考慮初始化的位置

可以發現當a的個數爲k的時候再添加一次b就結束算法了
所以考慮初始化dp[k][x]

PA = pa/(pa+pb) , PB = pb/(pa+pb)
當a的個數爲k的時候
假設第一次操作就放b那麼ab（子序列）的個數是k+x
第二次才放b那ab的個數是k+x+1
……
之後的第i個a產生貢獻的概率是PA^i

那麼總的期望就是x+k+sigma(PA^i)

用等比數列求和公式就可以算出此時的期望是x+k+pa/pb

當a爲1時後面纔可能存在ab子序列所以答案是dp[1][0]

然後隨便dp一下

#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <queue>
#include <cstdio>
#include <map>
#include <set>
#include <utility>
#include <stack>
#include <cstring>
#include <cmath>
#include <vector>
#include <ctime>
#include <bitset>
using namespace std;
#define pb push_back
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d%d",&n,&m)
#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define sld(n) scanf("%lld",&n)
#define sldd(n,m) scanf("%lld%lld",&n,&m)
#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)
#define sf(n) scanf("%lf",&n)
#define sff(n,m) scanf("%lf%lf",&n,&m)
#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)
#define ss(str) scanf("%s",str)
#define ansn() printf("%d\n",ans)
#define lansn() printf("%lld\n",ans)
#define r0(i,n) for(int i=0;i<(n);++i)
#define r1(i,e) for(int i=1;i<=e;++i)
#define rn(i,e) for(int i=e;i>=1;--i)
#define mst(abc,bca) memset(abc,bca,sizeof abc)
#define lowbit(a) (a&(-a))
#define all(a) a.begin(),a.end()
#define pii pair<int,int>
#define pll pair<long long,long long>
#define mp(aa,bb) make_pair(aa,bb)
#define lrt rt<<1
#define rrt rt<<1|1
#define X first
#define Y second
#define PI (acos(-1.0))
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1000000007 ;
const double eps=1e-9;
const int inf=0x3f3f3f3f;
//const ll infl = 10000000000000000;
const int maxn=  1e3+10;
const int maxm = 2e5+10;
//Pretests passed
int in(int &ret)
{
    char c;
    int sgn ;
    if(c=getchar(),c==EOF)return -1;
    while(c!='-'&&(c<'0'||c>'9'))c=getchar();
    sgn = (c=='-')?-1:1;
    ret = (c=='-')?0:(c-'0');
    while(c=getchar(),c>='0'&&c<='9')ret = ret*10+(c-'0');
    ret *=sgn;
    return 1;
}
ll qpow(ll x,ll k)
{
    ll res = 1;
    while(k)
    {
        if(k&1)res = res * x % mod;
        x = x* x %mod;
        k>>=1;
    }
    return res;
}
ll inv(ll x)
{
    return qpow(x,mod-2);
}
ll dp[maxn][maxn];int k,pa,pb;
ll cal(int a,int b)
{
    if(b>=k)return b;
    return dp[a][b];
}
int main()
{
#ifdef LOCAL
    freopen("input.txt","r",stdin);
//    freopen("output.txt","w",stdout);
#endif // LOCAL

//    int n,pa,pb;
    sddd(k,pa,pb);
    for(int i=0;i<k;++i)dp[k][i] = (i+k+pa*inv(pb))%mod;
    for(int i=k-1;i>=1;--i)
    {
        for(int j=k-1;j>=0;--j)
        {
            dp[i][j] = (pa*cal(i+1,j) + pb*cal(i,j+i)) % mod;
            dp[i][j] = (dp[i][j]*inv(pa+pb))%mod;
        }
    }
    ll ans = dp[1][0];
    lansn();
    return 0;
}