【AGC025B】RGB Color

题面描述

Takahashi has a tower which is divided into $N$ layers. Initially, all the layers are uncolored. Takahashi is going to paint some of the layers in red, green or blue to make a beautiful tower. He defines the beauty of the tower as follows:

The beauty of the tower is the sum of the scores of the $N$ layers, where the score of a layer is $A$ if the layer is painted red, $A+B$ if the layer is painted green, $B$ if the layer is painted blue, and 0 if the layer is uncolored.

Here, $A$ and $B$ are positive integer constants given beforehand. Also note that a layer may not be painted in two or more colors.

Takahashi is planning to paint the tower so that the beauty of the tower becomes exactly $K$. How many such ways are there to paint the tower? Find the count modulo 998244353. Two ways to paint the tower are considered different when there exists a layer that is painted in different colors, or a layer that is painted in some color in one of the ways and not in the other.

翻译

你有 $n$ 个格子排成一排，每个格子可以涂成红、蓝、绿或不涂色，得分分别为 $A$ , $B$ , $A + B$ , $0$ 。求使总得分为 $K$ 的方案数，答案对 $998244353$ 取模

思路

其实感觉主要是翻译的锅导致大家做不出来。

注意到题面中的信息 A+B if the layer is painted green

为什么用 $A+B$ ?

因为绿色是红色加蓝色。

这提示了我们将绿色看为红色和蓝色都填。

那么题意就变成了 $n$ 个格子，每个格子填红、蓝或者都填或者都都不填。

那么我们就可以用组合数随便算了。

枚举红色填的个数（包含两种颜色都填），可以直接算出蓝色的个数。

$\tbinom{n}{cnt} * \tbinom{n}{\frac{k - cnt * a}{b}}$

代码

/*

* @Copyright: Copyright  2021 昕橘玥

* @Powered: Powered by .NET 5.0 on Kubernetes

* @Author: JuyueXin.

* @Date:   2021-09-17 18:20:28

* @Email: 8950466@qq.com

* @Last Modified by:   JuyueXin.

* @Last Modified time: 2021-09-17 18:57:07

*/

#include <bits/stdc++.h>

using namespace std;

#define int long long

int read(int x = 0, bool f = false, char ch = getchar()) {

	for (; !isdigit(ch); ch = getchar()) f |= (ch == '-');

	for (; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + (ch ^ 48);

	return f ? ~x + 1 : x;

}

const int mod = 998244353, N = 3e5 + 5;

int n, m, ans, A, B;

int fac[N], inv[N];

int qpow(int x, int y) {

	int ans = 1;

	for (; y; y >>= 1, x = (1ll * x * x) % mod) if (y & 1) ans = (1ll * ans * x) % mod;

	return ans;

}

int C(int x, int y) {

	if (x < y) return 0;

	return 1ll * fac[x] * inv[y] % mod * inv[x - y] % mod;

}

signed main() {

	n = read(), A = read(), B = read(), m = read(); fac[0] = 1;

	for (int i = 1; i <= n; ++i) fac[i] = (1ll * fac[i - 1] * i) % mod;

	inv[n] = qpow(fac[n], mod - 2);

	for (int i = n - 1; ~i; --i) inv[i] = (1ll * inv[i + 1] * (i + 1)) % mod;

	for (int i = 0; i <= n; ++i) {

		if (m < i * A) break;

		if (!((m - i * A) % B)) ans = (1ll * ans + 1ll * C(n, i) * C(n, (m - i * A) / B) % mod) % mod;

	} return printf("%lld\n", ans), 0;

}