CodeForces 668C Little Artem and Random Variable（数学）

题意：有两个骰子，每个骰子有n面，现在你需要求每个骰子扔到每一面的概率是多少，现在给你扔到min(a,b)=i的概率和max(a,b)=i的概率。

思路：由题意知道P(max(a,b))=i P(min(a,b))=i

    那么显然可以推导出来

    P(max(a,b)<=i) = P(a<=i)*P(b<=i)

    P(min(a,b)>=i) = P(a>=i)*P(b>=i) = (1-P(a<=i-1))*(1-P(b<=i-1))

    上面那个式子可以推导出P(a<=i)+P(b<=i)=1+P(a<=i)*P(b<=i)-P(min(a,b)>=i+1)

    同时P(a=i) = P(a<=i) - P(a<=i-1)，然后就是记录Pmax的前缀和和Pmin的后缀和解方程就可以了

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6+7;
double prea[maxn],preb[maxn],a[maxn],b[maxn];
int main()
{
     int n;
	 scanf("%d",&n);
	 for (int i = 1;i<=n;i++)
		 scanf("%lf",&prea[i]);
	 for (int i = 1;i<=n;i++)
		 scanf("%lf",&preb[i]);
	 for (int i = 1;i<=n;i++)
		 prea[i]+=prea[i-1];
	 for (int i = n;i;i--)
		 preb[i]+=preb[i+1];
	 for (int i = 1;i<=n;i++)
	 {
		 double aa = 1;
		 double bb = -(1+prea[i]-preb[i+1]);
		 double cc = prea[i];
		 double delta = max(bb*bb-4*aa*cc,0.0);
		 a[i] = (-bb-sqrt(delta)) / (2*aa);
		 b[i] = (-bb+sqrt(delta)) / (2*aa);
	 }
	 for (int i = 1;i<=n;i++)
		 printf("%lf ",a[i]-a[i-1]);
	 printf("\n");
	 for (int i = 1;i<=n;i++)
		 printf("%lf ",b[i]-b[i-1]);
	 printf("\n");
}

Description

Little Artyom decided to study probability theory. He found a book with a lot of nice exercises and now wants you to help him with one of them.

Consider two dices. When thrown each dice shows some integer from 1 to n inclusive. For each dice the probability of each outcome is given (of course, their sum is 1), and different dices may have different probability distributions.

We throw both dices simultaneously and then calculate values max(a, b) and min(a, b), where a is equal to the outcome of the first dice, while b is equal to the outcome of the second dice. You don't know the probability distributions for particular values on each dice, but you know the probability distributions for max(a, b) and min(a, b). That is, for each x from 1 to n you know the probability thatmax(a, b) would be equal to x and the probability that min(a, b) would be equal to x. Find any valid probability distribution for values on the dices. It's guaranteed that the input data is consistent, that is, at least one solution exists.

Input

First line contains the integer n (1 ≤ n ≤ 100 000) — the number of different values for both dices.

Second line contains an array consisting of n real values with up to 8 digits after the decimal point  — probability distribution formax(a, b), the i-th of these values equals to the probability that max(a, b) = i. It's guaranteed that the sum of these values for one dice is 1. The third line contains the description of the distribution min(a, b) in the same format.

Output

Output two descriptions of the probability distribution for a on the first line and for b on the second line.

The answer will be considered correct if each value of max(a, b) and min(a, b) probability distribution values does not differ by more than 10^- 6 from ones given in input. Also, probabilities should be non-negative and their sums should differ from 1 by no more than10^- 6.

Sample Input

Input

2
0.25 0.75
0.75 0.25

Output

0.5 0.5 
0.5 0.5 

Input

3
0.125 0.25 0.625
0.625 0.25 0.125

Output

0.25 0.25 0.5 
0.5 0.25 0.25