HDU-6397(2018 Multi-University Training Contest 8) Character Encoding(生成函数+组合数学)

题意

从$0$到$n-1$的数字里可重复的取至多$m$个数的和等于$k$的方案数。

思路

显然的生成函数的思路为构造

$(1+x+x^{2}+...+x^{n-1})^{m}$

那么$x^{k}$的系数即答案。等比数列求和后得到

$ \frac {(1-x^n)^m} {(1-x)^m}$

对分子二项式展开得到

$(1-x^n)^m = \sum_{i=0}^m C_m^{i}(-1)^i * x^{n*i}$

对分母根据泰勒展开得到

$(1-x)^{-m} = \sum_{j = 0}^{\infty }C_{m+j-1}^{j}x^{j}$

代码

#include <bits/stdc++.h>
#define DBG(x) cerr << #x << " = " << x << endl;

using namespace std;

const int N = 200000 + 5;
const int mod = 998244353;
typedef long long LL;

int t, n, m, k;
LL fac[N], invf[N];

LL add(LL a, LL b) {
    LL res = a + b;
    if(res < 0) res += mod;
    return res % mod;
}

LL mul(LL a, LL b) {
    LL res = a * b;
    if(res < 0) res += mod;
    return res % mod;
}

LL qpow(LL a, LL b) {
    LL res = 1;
    while(b) {
        if(b & 1) res = mul(res, a);
        a = mul(a, a);
        b /= 2;
    }
    return res;
}

void init() {
    fac[0] = fac[1] = 1;
    for(int i = 2; i < N; i++) fac[i] = mul(fac[i - 1], 1LL * i);
    invf[N - 1] = qpow(fac[N - 1], mod - 2);
    for(int i = N - 2; i >= 0; i--) invf[i] = mul(invf[i + 1], 1LL * i + 1);
}

LL C(int n, int m) {
    if(n < 0 || m < 0 || n < m) return 0;
    return mul(fac[n], mul(invf[m], invf[n - m]));
}

int main() {
    init();
    scanf("%d", &t);
    while(t--) {
        LL ans = 0;
        scanf("%d%d%d", &n, &m, &k);
        for(LL i = 0; k - n * i >= 0; i++) {
            LL j = k - n * i;
            LL tmp = (i % 2 == 0 ? 1 : -1);
            ans = add(ans, mul(tmp, mul(C(m, i), C(m + j - 1, m - 1))));
        }
        printf("%lld\n", ans % mod);
    }
    return 0;
}