51nod 1341 混合序列

把 \(a_ib_{n-i}\) 看成一项求和
则 \(a_{i+1}b_{n-i-1}=(a_i * p + r)*3*q^{n-i-1}=a_i*3*q^{n-i}*(\frac{p}{q}) + 3*r*q^{n-i}*(\frac{1}{q})=a_ib_{n-i}*(\frac{p}{q})+ 3*r*q^{n-i}*(\frac{1}{q})\)
然后就可以构造矩阵加一维求和了

#include <bits/stdc++.h>
#define pb push_back
#define fi first
#define se second
#define pii pair<int, int>
#define pli pair<ll, int>
#define lp p << 1
#define rp p << 1 | 1
#define mid ((l + r) / 2)
#define lowbit(i) ((i) & (-i))
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
#define rep(i,a,b) for(int i=(a);i<(b);i++)
#define per(i,a,b) for(int i=((b)-1);i>=(a);i--)
#define Edg int ccnt=1,head[N],to[E],ne[E];void addd(int u,int v){to[++ccnt]=v;ne[ccnt]=head[u];head[u]=ccnt;}void add(int u,int v){addd(u,v);addd(v,u);}
#define Edgc int ccnt=1,head[N],to[E],ne[E],c[E];void addd(int u,int v,int w){to[++ccnt]=v;ne[ccnt]=head[u];c[ccnt]=w;head[u]=ccnt;}void add(int u,int v,int w){addd(u,v,w);addd(v,u,w);}
#define es(u,i,v) for(int i=head[u],v=to[i];i;i=ne[i],v=to[i])
const int MOD = 1e9 + 7;
void M(int &x) {if (x >= MOD)x -= MOD; if (x < 0)x += MOD;}
int qp(int a, int b = MOD - 2) {int ans = 1; for (; b; a = 1LL * a * a % MOD, b >>= 1)if (b & 1)ans = 1LL * ans * a % MOD; return ans % MOD;}
template<class T>T gcd(T a, T b) { while (b) { a %= b; std::swap(a, b); } return a; }
template<class T>bool chkmin(T &a, T b) { return a > b ? a = b, 1 : 0; }
template<class T>bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
char buf[1 << 21], *p1 = buf, *p2 = buf;
inline char getc() {
    return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;
}
inline int _() {
    int x = 0, f = 1; char ch = getc();
    while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getc(); }
    while (ch >= '0' && ch <= '9') { x = x * 10ll + ch - 48; ch = getc(); }
    return x * f;
}

int p, q, r, n;
struct Mat {
    int mat[4][4];
    Mat() { memset(mat, 0, sizeof(mat)); }
    Mat operator * (const Mat &p) const {
        Mat c;
        rep (i, 0, 3) {
            rep (j, 0, 3) {
                ll temp = 0;
                rep (k, 0, 3) {
                    temp += 1ll * mat[i][k] * p.mat[k][j];
                    if (temp >= (1ll << 60)) temp %= MOD;
                }
                if (temp >= MOD) temp %= MOD;
                c.mat[i][j] = temp;
            }
        }
        return c;
    }
    void print() {
        rep (i, 0, 3) {
            rep (j, 0, 3) {
                printf("%d%c", mat[i][j], " \n"[j == 2]);
            }
        }
    }
};
Mat qp(Mat a, int b) {
    Mat res;
    rep(i, 0, 3) res.mat[i][i] = 1;
    while (b) {
        if (b & 1) res = res * a;
        a = a * a;
        b >>= 1;
    }
    return res;
}

int main() {
#ifdef LOCAL
    freopen("ans.out", "w", stdout);
#endif
    p = _(), q = _(), r = _(), n = _();
    Mat base;
    base.mat[0][0] = ll(p) * qp(q) % MOD;
    base.mat[1][0] = base.mat[1][1] = qp(q);
    base.mat[0][2] = base.mat[2][2] = 1;
    base = qp(base, n + 1);
    Mat res;
    res.mat[0][0] = 0; res.mat[0][1] = 3ll * r * qp(q, n) % MOD;
    res = res * base;
    printf("%d\n", res.mat[0][2]);
#ifdef LOCAL
    printf("%.10f\n", (db)clock() / CLOCKS_PER_SEC);
#endif
    return 0;
}