#138. 类欧几里得算法

​​#138. 类欧几里得算法​​

以下除法均为向下取整， 定义 f ( a , b , c , n , k 1 , k 2 ) = ∑ x = 0 n x k 1 ( a × x + b c ) k 2 f(a, b, c, n, k_1, k_2) = \sum\limits_{x = 0} ^{n} x ^{k_1} \left(\frac{a \times x + b}{c}\right) ^ {k_2} f(a,b,c,n,k1​,k2​)=x=0∑n​xk1​(ca×x+b​)k2​。

∑ x = 0 n x k 1 ( a × x + b c ) k 2 , ( 1 ≤ n , a , c ≤ 1 0 9 , 0 ≤ b ≤ 1 0 9 , 0 ≤ k 1 + k 2 ≤ 10 ) \sum_{x = 0} ^{n} x ^{k_1} \left(\frac{a \times x + b}{c}\right) ^ {k_2}, (1 \le n, a, c \le 10 ^ 9, 0 \le b \le 10 ^ 9, 0 \le k_1 + k_2 \le 10)\\ x=0∑n​xk1​(ca×x+b​)k2​,(1≤n,a,c≤109,0≤b≤109,0≤k1​+k2​≤10)

a ≥ c , a ′ = a   %   c , k = a c a \ge c, a' = a\ \%\ c, k = \frac{a}{c} a≥c,a′=a % c,k=ca​：

∑ x = 0 n x k 1 ( k x + a ′ × x + b c ) k 2 ∑ x = 0 n x k 1 ∑ i = 0 k 2 C k 2 i ( k x ) i ( a ′ × x + b c ) k 2 − i ∑ i = 0 k 2 ( C k 2 i k i ) ∑ x = 0 n x i + k 1 ( a ′ × x + b c ) k 2 − i f ( a , b , c , n , k 1 , k 2 ) = ∑ i = 0 k 2 C k 2 i k i f ( a ′ , b , c , n , i + k 1 , k 2 − i ) \sum_{x = 0} ^{n} x ^{k_1} \left(kx + \frac{a' \times x + b}{c} \right) ^ {k_2}\\ \sum_{x = 0} ^{n} x ^{k_1}\sum_{i = 0} ^{k_2} C_{k_2} ^{i} (kx) ^ i \left(\frac{a' \times x + b}{c} \right) ^{k_2 - i}\\ \sum_{i = 0} ^{k_2} \left(C_{k_2} ^{i} k ^{i}\right) \sum_{x = 0} ^{n} x ^{i + k_1} \left( \frac{a' \times x + b}{c} \right) ^{k_2 - i}\\ f(a, b, c, n, k_1, k_2) = \sum_{i = 0} ^{k_2} C_{k_2} ^{i} k ^{i} f(a', b, c, n, i + k_1, k_2 - i)\\ x=0∑n​xk1​(kx+ca′×x+b​)k2​x=0∑n​xk1​i=0∑k2​​Ck2​i​(kx)i(ca′×x+b​)k2​−ii=0∑k2​​(Ck2​i​ki)x=0∑n​xi+k1​(ca′×x+b​)k2​−if(a,b,c,n,k1​,k2​)=i=0∑k2​​Ck2​i​kif(a′,b,c,n,i+k1​,k2​−i)

b ≥ c , b ′ = b   %   c , k = b c b \ge c, b' = b \ \%\ c, k = \frac{b}{c} b≥c,b′=b % c,k=cb​：

∑ x = 0 n x k 1 ( a × x + b ′ c + k ) k 2 ∑ x = 0 n x k 1 ∑ i = 0 k 2 C k 2 i k i ( a × x + b ′ c ) k 2 − i ∑ i = 0 k 2 C k 2 i k i ∑ x = 0 n x k 1 ( a × x + b ′ c ) k 2 − i f ( a , b , c , n , k 1 , k 2 ) = ∑ i = 0 k 2 C k 2 i k i f ( a , b ′ , c , n , k 1 , k 2 − i ) \sum_{x = 0} ^{n} x ^{k_1} \left(\frac{a \times x + b'}{c} + k \right) ^{k_2}\\ \sum_{x = 0} ^{n} x ^{k_1} \sum_{i = 0} ^{k_2} C_{k_2} ^{i} k ^{i} \left( \frac{a \times x + b'}{c} \right) ^{k_2 - i}\\ \sum_{i = 0} ^{k_2} C_{k_2} ^{i} k^{i} \sum_{x = 0} ^{n} x ^{k_1} \left(\frac{a \times x + b'}{c} \right) ^{k_2 - i}\\ f(a, b, c, n, k_1, k_2) = \sum_{i = 0} ^{k_2} C_{k_2} ^{i} k ^{i} f(a, b', c, n, k_1, k_2 - i)\\ x=0∑n​xk1​(ca×x+b′​+k)k2​x=0∑n​xk1​i=0∑k2​​Ck2​i​ki(ca×x+b′​)k2​−ii=0∑k2​​Ck2​i​kix=0∑n​xk1​(ca×x+b′​)k2​−if(a,b,c,n,k1​,k2​)=i=0∑k2​​Ck2​i​kif(a,b′,c,n,k1​,k2​−i)

a < c ,   b < c , m = a × n + b c a < c , \ b < c, m = \frac{a \times n + b}{c} a<c, b<c,m=ca×n+b​：

n k = ∑ i = 1 n i k − ( i − 1 ) k ∑ x = 0 n x k 1 ( a × x + b c ) k 2 ∑ x = 0 n x k 1 ∑ i = 1 m ( i k 2 − ( i − 1 ) k 2 ) [ a × x + b c ≥ i ] ∑ i = 0 m − 1 ( ( i + 1 ) k 2 − i k 2 ) ∑ x = 0 n x k 1 [ a × x + b c ≥ i + 1 ] ∑ i = 0 m − 1 ( ( i + 1 ) k 2 − i k 2 ) ∑ x = 0 n x k 1 − ∑ i = 0 m − 1 ( ( i + 1 ) k 2 − i k 2 ) ∑ x = 0 c × i + c − b − 1 a x k 1 n ^{k} = \sum_{i = 1} ^{n} i ^{k} - (i - 1) ^{k}\\ \sum_{x = 0} ^{n} x ^{k_1} \left(\frac{a \times x + b}{c}\right) ^ {k_2}\\ \sum_{x = 0} ^{n} x ^{k_1} \sum_{i = 1} ^{m} \left(i ^{k_2} - (i - 1) ^{k_2}\right)[\frac{a \times x + b}{c} \ge i]\\ \sum_{i = 0} ^{m - 1} \left((i + 1) ^{k_2} - i ^{k_2} \right) \sum_{x = 0} ^{n} x^{k_1} [\frac{a \times x + b}{c} \ge i + 1]\\ \sum_{i = 0} ^{m - 1} \left((i + 1) ^{k_2} - i ^{k_2} \right) \sum_{x = 0} ^{n} x^{k_1} - \sum_{i = 0} ^{m - 1} \left((i + 1) ^{k_2} - i ^{k_2} \right) \sum_{x = 0} ^{\frac{c \times i + c - b - 1}{a}} x ^{k_1}\\ nk=i=1∑n​ik−(i−1)kx=0∑n​xk1​(ca×x+b​)k2​x=0∑n​xk1​i=1∑m​(ik2​−(i−1)k2​)[ca×x+b​≥i]i=0∑m−1​((i+1)k2​−ik2​)x=0∑n​xk1​[ca×x+b​≥i+1]i=0∑m−1​((i+1)k2​−ik2​)x=0∑n​xk1​−i=0∑m−1​((i+1)k2​−ik2​)x=0∑ac×i+c−b−1​​xk1​

前项可以通过插值得到，我们考虑后项求和，

不难得到 ( i + 1 ) k 2 − i k 2 (i + 1) ^{k_2} - i ^{k_2} (i+1)k2​−ik2​，是一个 k 2 − 1 k_2 - 1 k2​−1次的多项式，我们可以通过插值预处理出其系数，设其为 A ( x ) A(x) A(x)。

对于 ∑ i = 0 c × i + c − b − 1 a x k 1 \sum\limits_{i = 0} ^{\frac{c \times i + c - b - 1}{a}} x ^{k_1} i=0∑ac×i+c−b−1​​xk1​，也是一个多项式，最高次幂为 k 1 + 1 k_1 + 1 k1​+1，设其为 B ( x ) B(x) B(x)。

∑ i = 0 m − 1 ( ( i + 1 ) k 2 − i k 2 ) ∑ x = 0 n x k 1 − ∑ x = 0 m − 1 ∑ i = 0 k 2 − 1 A i x i ∑ j = 0 k 1 + 1 B j ( c × x + c − b − 1 a ) j ∑ i = 0 m − 1 ( ( i + 1 ) k 2 − i k 2 ) ∑ x = 0 n x k 1 − ∑ i = 0 k 2 − 1 A i ∑ j = 0 k 1 + 1 B j ∑ x = 0 m − 1 x i ( c × x + c − b − 1 a ) j f ( a , b , c , n , k 1 , k 2 ) = ∑ i = 0 m − 1 ( ( i + 1 ) k 2 − i k 2 ) ∑ x = 0 n x k 1 − ∑ i = 0 k 2 − 1 ∑ j = 0 k 1 + 1 A i × B j × f ( a ′ , b ′ , c ′ , m − 1 , i , j ) \sum_{i = 0} ^{m - 1} \left((i + 1) ^{k_2} - i ^{k_2} \right) \sum_{x = 0} ^{n} x^{k_1} - \sum_{x = 0} ^{m - 1} \sum_{i = 0} ^{k_2 - 1} A_i x ^{i} \sum_{j = 0} ^{k_1 + 1} B_j \left( \frac{c \times x + c - b - 1}{a} \right) ^{j}\\ \sum_{i = 0} ^{m - 1} \left((i + 1) ^{k_2} - i ^{k_2} \right) \sum_{x = 0} ^{n} x^{k_1} - \sum_{i = 0} ^{k_2 - 1} A_i \sum_{j = 0} ^{k_1 + 1} B_j \sum_{x = 0} ^{m - 1} x ^{i} \left( \frac{c \times x + c - b - 1}{a} \right) ^{j}\\ f(a, b, c, n, k_1, k_2) = \sum_{i = 0} ^{m - 1} \left((i + 1) ^{k_2} - i ^{k_2} \right) \sum_{x = 0} ^{n} x^{k_1} - \sum_{i = 0} ^{k_2 - 1} \sum_{j = 0} ^{k_1 + 1} A_i \times B_j \times f(a', b', c', m - 1, i, j)\\ i=0∑m−1​((i+1)k2​−ik2​)x=0∑n​xk1​−x=0∑m−1​i=0∑k2​−1​Ai​xij=0∑k1​+1​Bj​(ac×x+c−b−1​)ji=0∑m−1​((i+1)k2​−ik2​)x=0∑n​xk1​−i=0∑k2​−1​Ai​j=0∑k1​+1​Bj​x=0∑m−1​xi(ac×x+c−b−1​)jf(a,b,c,n,k1​,k2​)=i=0∑m−1​((i+1)k2​−ik2​)x=0∑n​xk1​−i=0∑k2​−1​j=0∑k1​+1​Ai​×Bj​×f(a′,b′,c′,m−1,i,j)

只要预处理各种插值系数，即可在 O ( T × k 4 log ⁡ n ) O(T \times k ^4 \log n) O(T×k4logn)的复杂度内求解。

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 15;
const int P = 1e9 + 7;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
x = 0; int f = 1;
char c = getchar();
for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
x *= f;
}
template <typename T> void write(T x) {
if (x < 0) x = -x, putchar('-');
if (x > 9) write(x / 10);
putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
write(x);
puts("");
}
struct info {
int a[MAXN][MAXN];
};
int sum[MAXN][MAXN];
int binom[MAXN][MAXN];
int power(int x, int y) {
if (y == 0) return 1;
int tmp = power(x, y / 2);
if (y % 2 == 0) return 1ll * tmp * tmp % P;
else return 1ll * tmp * tmp % P * x % P;
}
void update(int &x, int y) {
x += y;
if (x >= P) x -= P;
}
info func(int n, int a, int b, int c) {
assert(n >= 0 && a >= 0 && b >= 0 && c >= 0);
info ans;
memset(ans.a, 0, sizeof(ans.a));
if (a == 0 || (1ll * a * n + b) / c == 0) {
for (int k1 = 0; k1 <= 10; k1++) {
int mul = 0, now = 1;
for (int i = 0; i <= k1 + 1; i++) {
update(mul, 1ll * now * sum[k1][i] % P);
now = 1ll * now * n % P;
      }
int base = (1ll * a * n + b) / c % P; now = 1;
for (int k2 = 0; k1 + k2 <= 10; k2++) {
ans.a[k1][k2] = 1ll * now * mul % P;
now = 1ll * now * base % P;
      }
    }
return ans;
  }
if (a >= c) {
info tmp = func(n, a % c, b, c);
for (int k1 = 0; k1 <= 10; k1++)
for (int k2 = 0; k1 + k2 <= 10; k2++) {
int now = 1, base = a / c;
for (int i = 0; i <= k2; i++) {
update(ans.a[k1][k2], 1ll * binom[k2][i] * now % P * tmp.a[k1 + i][k2 - i] % P);
now = 1ll * now * base % P;
      }
    }
return ans;
  }
if (b >= c) {
info tmp = func(n, a, b % c, c);
for (int k1 = 0; k1 <= 10; k1++)
for (int k2 = 0; k1 + k2 <= 10; k2++) {
int now = 1, base = b / c;
for (int i = 0; i <= k2; i++) {
update(ans.a[k1][k2], 1ll * binom[k2][i] * now % P * tmp.a[k1][k2 - i] % P);
now = 1ll * now * base % P;
      }
    }
return ans;
  }
int m = (1ll * a * n + b) / c;
info tmp = func(m - 1, c, c - b - 1, a);
for (int k1 = 0; k1 <= 10; k1++) {
int all = 0, now = 1;
for (int i = 0; i <= k1 + 1; i++) {
update(all, 1ll * now * sum[k1][i] % P);
now = 1ll * now * n % P;
    }
for (int k2 = 0; k1 + k2 <= 10; k2++) {
ans.a[k1][k2] = 1ll * power(m, k2) * all % P;
for (int i = 0; i <= k2 - 1; i++)
for (int j = 0; j <= k1 + 1; j++) {
int coef = 1ll * binom[k2][i] * sum[k1][j] % P;
update(ans.a[k1][k2], P - 1ll * coef * tmp.a[i][j] % P);
      }
    }
  }
return ans;
}
void Lagrange(int n, int *x, int *y, int *a) {
static int p[MAXN], q[MAXN];
memset(p, 0, sizeof(p)); p[0] = 1;
for (int i = 1; i <= n; i++) {
for (int j = i - 1; j >= 0; j--) {
p[j + 1] = (p[j + 1] + p[j]) % P;
p[j] = (P - 1ll * p[j] * x[i] % P) % P;
    }
  }
for (int i = 1; i <= n; i++) {
memset(q, 0, sizeof(q));
for (int j = n - 1; j >= 0; j--)
q[j] = (p[j + 1] + 1ll * q[j + 1] * x[i]) % P;
int now = 1;
for (int j = 1; j <= n; j++)
if (j != i) now = 1ll * now * (x[i] - x[j]) % P;
now = power((P + now) % P, P - 2);
for (int j = 0; j <= n; j++)
q[j] = 1ll * q[j] * now % P;
for (int j = 0; j <= n; j++)
a[j] = (a[j] + 1ll * q[j] * y[i]) % P;
  }
}
int main() {
for (int i = 0; i <= 10; i++) {
static int pos[MAXN], now[MAXN];
now[0] = power(0, i);
for (int j = 1; j <= i + 2; j++) {
now[j] = (now[j - 1] + power(j, i)) % P;
pos[j] = j;
    }
Lagrange(i + 2, pos, now, sum[i]);
  }
for (int i = 0; i <= 10; i++) {
binom[i][0] = 1;
for (int j = 1; j <= i; j++)
binom[i][j] = binom[i - 1][j - 1] + binom[i - 1][j];
  }
int T; read(T);
while (T--) {
int n, a, b, c, k1, k2;
read(n), read(a), read(b), read(c), read(k1), read(k2);
writeln(func(n, a, b, c).a[k1][k2]);
  }
return 0;
}