【AGC025B】RGB Color
【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;
}