【AGC025B】RGB Color

甜桃奶芙發表於2021-09-17
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【AGC025B】RGB Color

題面描述

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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.

翻譯

你有 \(n\) 個格子排成一排，每個格子可以塗成紅、藍、綠或不塗色，得分分別為 \(A\) , \(B\) , \(A + B\) , \(0\) 。求使總得分為 \(K\) 的方案數，答案對 \(998244353\) 取模

思路

其實感覺主要是翻譯的鍋導致大家做不出來。

注意到題面中的資訊 A+B if the layer is painted green

為什麼用 \(A+B\) ?

因為綠色是紅色加藍色。

這提示了我們將綠色看為紅色和藍色都填。

那麼題意就變成了 \(n\) 個格子，每個格子填紅、藍或者都填或者都都不填。

那麼我們就可以用組合數隨便算了。

列舉紅色填的個數（包含兩種顏色都填），可以直接算出藍色的個數。

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

程式碼

/*
* @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;
}

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