zoukankan      html  css  js  c++  java
  • AGC 025 B

    B - RGB Coloring


    Time limit : 2sec / Memory limit : 1024MB

    Score : 700 points

    Problem Statement

    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.

    Constraints

    • 1≤N≤3×105
    • 1≤A,B≤3×105
    • 0≤K≤18×1010
    • All values in the input are integers.

    Input

    Input is given from Standard Input in the following format:

    N A B K
    

    Output

    Print the number of the ways to paint tiles, modulo 998244353.


    Sample Input 1

    Copy
    4 1 2 5
    

    Sample Output 1

    Copy
    40
    

    In this case, a red layer worth 1 points, a green layer worth 3 points and the blue layer worth 2 points. The beauty of the tower is 5 when we have one of the following sets of painted layers:

    • 1 green, 1 blue
    • 1 red, 2 blues
    • 2 reds, 1 green
    • 3 reds, 1 blue

    The total number of the ways to produce them is 40.


    Sample Input 2

    Copy
    2 5 6 0
    

    Sample Output 2

    Copy
    1
    

    The beauty of the tower is 0 only when all the layers are uncolored. Thus, the answer is 1.


    Sample Input 3

    Copy
    90081 33447 90629 6391049189
    

    Sample Output 3

    Copy
    577742975


    pkusc之后非常难过qwq,刷道水题安慰自己qwq。
    (这可能是组合计数模板题??)

    #include<iostream>
    #include<cstdio>
    #include<cstdlib>
    #include<algorithm>
    #include<cstring>
    #include<ctime>
    #define ll long long
    using namespace std;
    const int maxn=300005,ha=998244353;
    inline void add(int &x,int y){ x+=y; if(x>=ha) x-=ha;}
    inline int ksm(int x,int y){
    	int an=1;
    	for(;y;y>>=1,x=x*(ll)x%ha) if(y&1) an=an*(ll)x%ha;
    	return an; 
    }
    
    int jc[maxn],ni[maxn];
    int N,A,B,ans;
    ll K;
    
    inline int C(int x,int y){ return jc[x]*(ll)ni[y]%ha*(ll)ni[x-y]%ha;}
    
    inline void init(){
    	jc[0]=1;
        for(int i=1;i<=N;i++) jc[i]=jc[i-1]*(ll)i%ha;
        ni[N]=ksm(jc[N],ha-2);
        for(int i=N;i;i--) ni[i-1]=ni[i]*(ll)i%ha;
    }
    
    inline void solve(){
    	const int b=B,T=min((ll)N,K/A);
    	
    	for(int i=0;i<=T;i++){
    		if((K-A*(ll)i)%b) continue;
    		ll lef=(K-A*(ll)i)/b;
    		if(lef>N) continue;
    		
    		add(ans,C(N,i)*(ll)C(N,lef)%ha);
    	}
    }
    
    int main(){
    	scanf("%d%d%d%lld",&N,&A,&B,&K);
    	
    	init(),solve();
    	
    	printf("%d
    ",ans);
    	return 0;
    }
    

      

  • 相关阅读:
    工坊第五天
    工坊第四天
    工坊第三天
    工坊第二天
    工坊第一天
    莫队 优雅暴力出奇迹
    状压 DP 总结
    关于MatlabGUI清除WorkSpace的用法
    ArduinoNano卡在上传,无法烧录
    两轮差速驱动机器人的坐标轨迹计算
  • 原文地址:https://www.cnblogs.com/JYYHH/p/9138191.html
Copyright © 2011-2022 走看看