华华跟奕奕玩游戏

 https://ac.nowcoder.com/acm/contest/894/C

思路：设a[i] 为第i次操作剩下的黑球个数；

反思：矩阵快速幂的b^step,写成了b^(step-1)，小数据看不出来。。。

 

 

#include<algorithm>
#include<set>
#include<queue>
#include<cmath>
#include<cstring>
#include<iostream>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstdio>
#include<map>
#include<stack>
#include<string>
#include<bits/stdc++.h>
using namespace std;

#define sfi(i) scanf("%d",&i)
#define pri(i) printf("%d\n",i)
#define sff(i) scanf("%lf",&i)
#define ll long long
#define ull unsigned long long
#define mem(x,y) memset(x,y,sizeof(x))
#define INF 0x3f3f3f3f
#define eps 1e-16
#define PI acos(-1)
#define lowbit(x) ((x)&(-x))
#define zero(x) (((x)>0?(x):-(x))<eps)
#define fl() printf("flag\n")
#define MOD(x) ((x%mod)+mod)%mod
#define endl '\n'
#define pb push_back`
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)

const int maxn=2e5+9;
const int mod=1e9+7;

inline ll read()
{
    ll f=1,x=0;

    char ss=getchar();
    while(ss<'0'||ss>'9')
    {
        if(ss=='-')f=-1;ss=getchar();
    }
    while(ss>='0'&&ss<='9')
    {
        x=x*10+ss-'0';ss=getchar();
    }    return f*x;
}

ll power(ll x,ll n)
{
    ll ans=1;
    while(n)
    {
        if(n&1) ans=ans*x%mod;
        n>>=1;
        x=x*x%mod;
    }
    return ans;
}

ll a,b,n,m,step;
struct M
{
    ll a[3][3];
    M()
    {
        mem(a,0);
    }

};
M Mmul(M x,M y)
{
    M res;
    for(int i=1;i<=2;i++)
    {
        for(int j=1;j<=2;j++)
        {
            for(int k=1;k<=2;k++)
            {
                res.a[i][j]=res.a[i][j]+(x.a[i][k]*y.a[k][j])%mod;
                res.a[i][j]%=mod;
            }
        }
    }
    return res;
}

M Mpower(M x,ll n)
{
    M res;
    for(int i=1;i<=2;i++)
    {
        res.a[i][i]=1;
    }
    while(n)
    {
        if(n&1)
        {
            res=Mmul(res,x);
        }
        x=Mmul(x,x);
        n>>=1;
    }
    return res;
}

int main()
{
    FAST_IO;
    //freopen("input.txt","r",stdin);

    cin>>n>>m>>step>>a>>b;

    ll inv=power(n+m+1,mod-2);

    ll k=(n+m)*inv%mod;

    ll invb=power(b,mod-2);

    ll c=a*(n+m)%mod;
    c=c*inv%mod;
    c=c*invb%mod;

    M tmp;
    tmp.a[1][1]=k;
    tmp.a[1][2]=1;
    tmp.a[2][1]=0;
    tmp.a[2][2]=1;

    M t;
    t=Mpower(tmp,step);

    M ans;
    ans.a[1][1]=n;
    ans.a[2][1]=c;

    ans=Mmul(t,ans);

    cout<<ans.a[1][1]<<endl;
    //cout<<(t.a[1][1]*n%mod+t.a[1][2]*c%mod)%mod<<endl;
    return 0;
}