【题目链接】

• 点击打开链接
  

【思路要点】

• 列出方程，高斯消元。
• 我们发现在消元时每一行非零的位置为\(O(1)\)的，我们只要处理这些位置即可。
• 时间复杂度\(O(TN^2)\)，需要卡常数。

【代码】

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1505;
const int P = 1e9 + 7;
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("");
}
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;
}
int n, m, k, p, hero[MAXN], c[MAXN];
int fac[MAXN], fbc[MAXN], inv[MAXN], a[MAXN][MAXN];
void gauss(int n) {
	for (int i = 1; i <= n; i++) {
		for (int j = i; j <= n; j++)
			if (a[j][i]) {
				if (i != j) swap(a[i], a[j]);
				break;
			}
		a[i][i] = (a[i][i] % P + P) % P;
		int inv = power(a[i][i], P - 2);
		for (int j = 1; j <= n + 1; j++)
			a[i][j] = 1ll * a[i][j] * inv % P;
		for (int j = min(i + 1, p); j <= n; j++) {
			if (i == j || a[j][i] == 0) continue;
			long long tmp = a[j][i];
			a[j][i] = (a[j][i] - tmp * a[i][i]) % P;
			if (i + 1 <= n) a[j][i + 1] = (a[j][i + 1] - tmp * a[i][i + 1]) % P;
			a[j][n + 1] = (a[j][n + 1] - tmp * a[i][n + 1]) % P;
		}
	}
}
int main() {
	fac[0] = 1;
	for (int i = 1; i < MAXN; i++)
		fac[i] = 1ll * fac[i - 1] * i % P;
	inv[MAXN - 1] = power(fac[MAXN - 1], P - 2);
	for (int i = MAXN - 2; i >= 0; i--)
		inv[i] = inv[i + 1] * (i + 1ll) % P;
	int T; read(T);
	while (T--) {
		read(n), read(p), read(m), read(k);
		if (k == 0 || k == 1 && m == 0) {
			printf("-1\n");
			continue;
		}
		fbc[0] = 1;
		for (int i = 1; i <= n; i++)
			fbc[i] = fbc[i - 1] * (k - i + 1ll) % P;
		memset(a, 0, sizeof(a));
		int tinv = power(m + 1, P - 2);
		hero[0] = power(1ll * tinv * m % P, k);
		int invm = power(m, P - 2);
		for (int i = 1; i <= n; i++)
			hero[i] = 1ll * hero[i - 1] * invm % P;
		if (m == 0 && k <= n) hero[k] = 1;
		for (int i = 0; i <= n; i++)
			c[i] = 1ll * fbc[i] * inv[i] % P * hero[i] % P;
		for (int i = 1; i <= n - 1; i++) {
			a[i][i] = P - 1; a[i][n + 1] = P - 1;
			int tmp = 1ll * tinv * m % P;
			for (int j = i; j >= 1; j--)
				a[i][j] = (a[i][j] + 1ll * tmp * c[i - j]) % P;
			if (m == 0) tmp = 1;
			else tmp = tinv;
			for (int j = i + 1; j >= 1; j--)
				a[i][j] = (a[i][j] + 1ll * tmp * c[i + 1 - j]) % P;
		}
		a[n][n] = P - 1, a[n][n + 1] = P - 1;
		for (int j = n; j >= 1; j--)
			a[n][j] = (a[n][j] + c[n - j]) % P;
		gauss(n);
		writeln((a[p][n + 1] % P + P) % P);
	}
	return 0;
}