首先普通 $DP$ 式子很好列

令 $f[i][j]$ 表示 $i$ 个小朋友分 $j$ 个糖果且每人至少一颗的答案，则有
$$
f[i][j] = \sum\limits_{k = 1}^j f[i - 1][j - k](Ok^2 + Sk + U)
$$
接下来用生成函数优化该式

令 $g[k] = Ok^2 + Sk + U$，则
$$
f_i(x) = f_{i - 1}(x)g(x)
$$
其中 $f_0(x) = 1$

所以答案即为
$$
[x^M] \sum\limits_{i = 0}^A f_i(x) = [x^M] \sum\limits_{i = 0}^A (g(x))^i = [x^M] \frac{1 - (g(x))^{A + 1}}{1 - g(x)}
$$

代码

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>

using namespace std;

typedef long long LL;

const int MAXN = 1e05 + 10;

const double PI = acos (- 1.0);

int N, M;
LL MOD;
LL O, S, U;
LL a[MAXN]= {0}, b[MAXN]= {0};

struct mcomplex {
	double a, b;

	mcomplex (double fa = 0.0, double fb = 0.0) :
		a (fa), b (fb) {}

	mcomplex operator + (const mcomplex& p) const {
		return mcomplex (a + p.a, b + p.b);
	}
	mcomplex operator - (const mcomplex& p) const {
		return mcomplex (a - p.a, b - p.b);
	}
	mcomplex operator * (const mcomplex& p) const {
		return mcomplex (a * p.a - b * p.b, a * p.b + b * p.a);
	}
} ;

int oppo[MAXN << 2];
int limit;
void FFT (mcomplex* a, int inv) {
	for (int i = 0; i < limit; i ++)
		if (i < oppo[i])
			swap (a[i], a[oppo[i]]);
	for (int mid = 1; mid < limit; mid <<= 1) {
		mcomplex omega = mcomplex (cos (PI / (double) mid), inv * sin (PI / (double) mid));
		for (int n = mid << 1, j = 0; j < limit; j += n) {
			mcomplex x = mcomplex (1.0, 0);
			for (int k = 0; k < mid; k ++, x = x * omega) {
				mcomplex a1 = a[j + k], xa2 = x * a[j + mid + k];;
				a[j + k] = a1 + xa2;
				a[j + k + mid] = a1 - xa2;
			}
		}
	}
}
void aclimit (int maxl) {
	int n, lim;
	for (n = 1, lim = 0; n <= maxl; n <<= 1, lim ++);
	for (int i = 0; i < n; i ++)
		oppo[i] = (oppo[i >> 1] >> 1) | ((i & 1) << (lim - 1));
	limit = n;
}
mcomplex a1[MAXN << 2], a2[MAXN << 2];
void multiply (LL* A, LL* B, int n, int m) {
	aclimit (n + m);
	for (int i = 0; i < limit; i ++)
		a1[i] = a2[i] = mcomplex ();
	for (int i = 0; i <= n; i ++)
		a1[i] = mcomplex ((double) A[i], 0);
	for (int i = 0; i <= m; i ++)
		a2[i] = mcomplex ((double) B[i], 0);
	FFT (a1, 1), FFT (a2, 1);
	for (int i = 0; i < limit; i ++)
		a1[i] = a1[i] * a2[i];
	FFT (a1, - 1);
	for (int i = 0; i <= n + m; i ++)
		A[i] = (LL) (a1[i].a / limit + 0.5) % MOD;
}
LL inv[MAXN << 2]= {0};
LL temp[MAXN << 2]= {0};
void inverse (int deg, LL* A, LL* B) {
	if (deg == 1) {
		B[0] = 1;
		return ;
	}
	inverse ((deg + 1) >> 1, A, B);
	aclimit (deg << 1);
	for (int i = 0; i < limit; i ++)
		temp[i] = 0;
	for (int i = 0; i < deg; i ++)
		temp[i] = A[i];
	multiply (temp, B, deg, deg);
	multiply (temp, B, deg, deg);
	for (int i = 0; i < deg; i ++)
		B[i] = (2 * B[i] % MOD - temp[i] + MOD) % MOD;
}
LL ans[MAXN << 2]= {0};
void power (LL* A, int p) {
	ans[0] = 1;
	while (p) {
		if (p & 1)
			multiply (ans, a, M, M);
		multiply (a, a, M, M);
		p >>= 1;
	}
}

int getnum () {
	int num = 0;
	char ch = getchar ();
	int isneg = 0;

	while (! isdigit (ch)) {
		if (ch == '-')
			isneg = 1;
		ch = getchar ();
	}
	while (isdigit (ch))
		num = (num << 3) + (num << 1) + ch - '0', ch = getchar ();

	return isneg ? - num : num;
}

int main () {
	M = getnum (), MOD = getnum (), N = getnum ();
	O = getnum (), S = getnum (), U = getnum ();
	a[0] = 0;
	for (int i = 1; i <= M; i ++)
		a[i] = (O * i * i + S * i + U) % MOD;
	for (int i = 1; i <= M; i ++)
		b[i] = (- a[i] + MOD) % MOD;
	b[0] = 1;
	power (ans, N + 1);
	for (int i = 0; i <= M; i ++)
		ans[i] = (- ans[i] + MOD) % MOD;
	ans[0] = (ans[0] + 1) % MOD;
	inverse (M + 1, b, inv);
	multiply (ans, inv, M, M);
	cout << ans[M] << endl;

	return 0;
}

/*
4 100
4
1
0
0
*/