题目描述：求
[sum_{x=0}^nlfloorfrac{px+r}{q}
floor^{k_1}x^{k_2} mod (10^9+7)
]
数据范围：数据组数 (T=1000)，(n,p,q,rle 10^9)，(k_1,k_2ge 0,k_1+k_2le 10)。
真正的类欧几里得，过于令人自闭了。
所以把刚学的万能欧几里得用一用，设Node结构体为 ((cnt_1,cnt_2,ans_{i,j}))，表示 ( ext{U,R}) 的个数，和 (k_1=i,k_2=j) 时的答案。所以若 (C=A+B)，则
[egin{aligned}
C.cnt_1&=A.cnt_1+B.cnt_1 \
C.cnt_2&=A.cnt_2+B.cnt_2 \
C.ans_{a,b}&=A.ans_{a,b}+sum_{i=0}^asum_{j=0}^binom{a}{i}inom{b}{j}A.cnt_1^iA.cnt_2^jB.ans_{a-i,b-j}
end{aligned}
]
其中第三个柿子把 ((y+A.cnt_1)^a(x+A.cnt_2)^b) 展开就可以得到。
再把那个模板原封不动地套一遍，时间复杂度 (O(225logmax{p,q}))。
#include<bits/stdc++.h>
#define Rint register int
#define MP make_pair
#define PB push_back
#define fi first
#define se second
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
const int mod = 1000000007;
template<typename T>
inline void read(T &x){
int ch = getchar(); x = 0; bool f = false;
for(;ch < '0' || ch > '9';ch = getchar()) f |= ch == '-';
for(;ch >= '0' && ch <= '9';ch = getchar()) x = x * 10 + ch - '0';
if(f) x = -x;
}
inline void qmo(int &x){x += (x >> 31) & mod;}
template<typename T>
inline bool chmax(T &a, const T &b){if(a < b) return a = b, 1; return 0;}
template<typename T>
inline bool chmin(T &a, const T &b){if(a > b) return a = b, 1; return 0;}
int T, C[11][11];
LL n, p, q, r, k1, k2;
struct Node {
LL cnt1, cnt2;
int ans[11][11];
Node(){cnt1 = cnt2 = 0; memset(ans, 0, sizeof ans);}
Node operator = (const Node &o){cnt1 = o.cnt1; cnt2 = o.cnt2; memcpy(ans, o.ans, sizeof ans); return *this;}
Node operator * (const Node &o) const {
Node res;
res.cnt1 = cnt1 + o.cnt1;
res.cnt2 = cnt2 + o.cnt2;
memcpy(res.ans, ans, sizeof res.ans);
int x = cnt1 % mod, y = cnt2 % mod;
for(Rint al = 0;al <= k1;++ al)
for(Rint be = 0;be <= k2;++ be)
for(Rint i = 0, t1 = 1;i <= al;++ i, t1 = (LL) t1 * x % mod)
for(Rint j = 0, t2 = 1;j <= be;++ j, t2 = (LL) t2 * y % mod)
qmo(res.ans[al][be] += (LL) C[al][i] * C[be][j] % mod * o.ans[al - i][be - j] % mod * t1 % mod * t2 % mod - mod);
return res;
}
};
template<typename T>
T ksm(T a, LL b){
T res;
for(;b;b >>= 1, a = a * a) if(b & 1) res = res * a;
return res;
}
Node calc(LL p, LL q, LL r, LL n, const Node &a, const Node &b){
LL m = (p * n + r) / q;
if(!m) return ksm(b, n);
if(p >= q) return calc(p % q, q, r, n, a, ksm(a, p / q) * b);
return ksm(b, (q - r - 1) / p) * a * calc(q, p, (q - r - 1) % p, m - 1, b, a) * ksm(b, n - (m * q - r - 1) / p);
}
inline void work(){
Node res, A, B; read(n); read(p); read(r); read(q); read(k2); read(k1);
A.cnt1 = 1; B.cnt2 = 1; for(Rint i = 0;i <= k2;++ i) B.ans[0][i] = 1;
res.cnt1 = r / q; for(Rint i = 0, t = 1;i <= k1;++ i, t = (LL) t * res.cnt1 % mod) res.ans[i][0] = t;
res = res * calc(p, q, r % q, n, A, B);
printf("%d
", res.ans[k1][k2]);
}
int main(){
read(T);
for(Rint i = 0;i <= 10;++ i){
C[i][0] = 1;
for(Rint j = 1;j <= i;++ j) qmo(C[i][j] = C[i - 1][j] + C[i - 1][j - 1] - mod);
}
while(T --) work();
}