zoukankan      html  css  js  c++  java
  • Codeforces 908 D New Year and Arbitrary Arrangement

    Discription

    You are given three integers kpa and pb.

    You will construct a sequence with the following algorithm: Initially, start with the empty sequence. Each second, you do the following. With probability pa / (pa + pb), add 'a' to the end of the sequence. Otherwise (with probability pb / (pa + pb)), add 'b' to the end of the sequence.

    You stop once there are at least k subsequences that form 'ab'. Determine the expected number of times 'ab' is a subsequence in the resulting sequence. It can be shown that this can be represented by P / Q, where P and Q are coprime integers, and . Print the value of .

    Input

    The first line will contain three integers integer k, pa, pb (1 ≤ k ≤ 1 000, 1 ≤ pa, pb ≤ 1 000 000).

    Output

    Print a single integer, the answer to the problem.

    Example

    Input
    1 1 1
    Output
    2
    Input
    3 1 4
    Output
    370000006

    Note

    The first sample, we will keep appending to our sequence until we get the subsequence 'ab' at least once. For instance, we get the sequence 'ab' with probability 1/4, 'bbab' with probability 1/16, and 'aab' with probability 1/8. Note, it's impossible for us to end with a sequence like 'aabab', since we would have stopped our algorithm once we had the prefix 'aab'.

    The expected amount of times that 'ab' will occur across all valid sequences is 2.

    For the second sample, the answer is equal to .

    设f[i][j]为有i对ab,并且已经有j个a的期望,转移很好写,f[i][j]= (pa/(pa+pb))*f[i][j+1] + (pb/(pa+pb))*f[i+j][j] 、

    但是可以发现的是如果要计算所有状态的话j显然可以无限大,,,比如全是a的序列。。。。

    但是还可以发现,当i+j>=k的时候,(pb/(pa+pb))*f[i+j][j] 其实就等于 (pb/(pa+pb))*(i+j)。

    这样我们等比数列错位相减一下(需要化简一大堆式子,在这就懒得写了),可以得到一个边界:f[i][j]=i+j +pa/pb    (i+j>=n)

    然后f[i][0]=f[i][1],这个带第一个转移的式子就可以得到。。。。。

    /*
        设f[i][j]为有i对ab,目前已经有了j个a的ab期望个数 
    	1.f[i][j]= pa/pb + i+j ,其中i+j>=n  (这个推个式子然后生成函数一下就OJBK了)
    	2.f[i][0]=f[i][1] (这个也是代换一下就好了)
    	3.其他情况下,f[i][j]= (pa/(pa+pb))*f[i][j+1] + (pb/(pa+pb))*f[i+j][j] 
    */
    #include<bits/stdc++.h>
    #define ll long long
    const int ha=1000000007;
    const int maxn=1005;
    int inv[2000005];
    int n,pa,pb;
    int f[2005][1005];
    
    inline void init(){
    	inv[1]=1;
    	for(int i=2;i<=2000000;i++) inv[i]=-inv[ha%i]*(ll)(ha/i)%ha+ha;
    }
    
    inline int add(int x,int y){
    	x+=y;
    	if(x>=ha) return x-ha;
    	else return x;
    }
    
    inline void dp(){
    	int base=(pa*(ll)inv[pb]+(ll)n)%ha;
    	int PA=pa*(ll)inv[pa+pb]%ha,PB=pb*(ll)inv[pa+pb]%ha;
    	for(int i=n-1;i>=0;i--){
    		for(int j=n-i;j<=n;j++) f[i][j]=add(base,j-n+i);
    		for(int j=n-i-1;j;j--) f[i][j]=add(f[i][j+1]*(ll)PA%ha,f[i+j][j]*(ll)PB%ha);
    		f[i][0]=f[i][1];
    	}
    }
    
    int main(){
    	init();
    	scanf("%d%d%d",&n,&pa,&pb);
    	dp();
    	printf("%d
    ",f[0][0]);
    	return 0;
    }
    

      

  • 相关阅读:
    qt中线程的使用方法
    QT中定时器的使用方法
    Common Lisp学习笔记(八)
    Common Lisp学习笔记(七)
    Common Lisp学习笔记(六)
    vim使用笔记
    Django学习笔记:urlresolvers
    python closures and decorators
    规范命名:变量名的力量
    eclipse openGL glut运行环境配置
  • 原文地址:https://www.cnblogs.com/JYYHH/p/8455806.html
Copyright © 2011-2022 走看看