Problem:
Score : 300 points

Problem Statement

You are given two integer sequences of length N: a_1,a_2,..,a_N and b_1,b_2,..,b_N.
Determine if we can repeat the following operation zero or more times so that the sequences a and b become equal.

Operation: Choose two integers i and j (possibly the same) between 1 and N (inclusive), then perform the following two actions simultaneously:

Add 2 to a_i.

Add 1 to b_j.

Constraints

1 ≤ N ≤ 10 000

0 ≤ a_i,b_i ≤ 10^9 (1 ≤ i ≤ N)

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 .. a_N
b_1 b_2 .. b_N

Output

If we can repeat the operation zero or more times so that the sequences a and b become equal, print Yes; otherwise, print No.

Sample Input 1

3
1 2 3
5 2 2

Sample Output 1

Yes

For example, we can perform three operations as follows to do our job:

First operation: i=1 and j=2. Now we have a = \{3,2,3\}, b = \{5,3,2\}.

Second operation: i=1 and j=2. Now we have a = \{5,2,3\}, b = \{5,4,2\}.

Third operation: i=2 and j=3. Now we have a = \{5,4,3\}, b = \{5,4,3\}.

Sample Input 2

5
3 1 4 1 5
2 7 1 8 2

Sample Output 2

No

Sample Input 3

5
2 7 1 8 2
3 1 4 1 5

Sample Output 3

No

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>

#define P pair
#define MP make_pair

typedef long long int llint ;

using namespace std;

vector<llint> niz1;
vector<llint> niz2;
llint d,n,x;

int main()
{
	cin>>n;
	for(int i=0;i<n;i++)
	{
		cin>>x;
		niz1.push_back(x);
	}
	for(int i=0;i<n;i++)
	{
		cin>>x;
		niz2.push_back(x);
	}
	while(niz1.size()>1)
	{
		llint a=niz1[0];
		llint b=niz2[0];
		if(a==b)
		{
			niz1.erase(niz1.begin());
			niz2.erase(niz2.begin());
			continue;
		}
		if(a>b)
		{
			llint podjeli=a-b;
			podjeli*=2;
			for(int i=1;i<=podjeli%(niz1.size()-1);i++)
			niz1[i]++;
			for(int i=1;i<niz1.size();i++)
				niz1[i]+=podjeli/(niz1.size()-1);
							
			niz1.erase(niz1.begin());
			niz2.erase(niz2.begin());
		}
		if(a<b)
		{
			d=(b-a);
			d+=d%2;
			niz1[0]+=d;
			d/=2;
			for(int i=1;i<=d%(niz2.size()-1);i++)
				niz2[i]++;
			for(int i=1;i<niz2.size();i++)
				niz2[1]+=d/(niz2.size()-1);
		}
		llint m=1e18;
		for(int i=0;i<niz1.size();i++)
		{
			m=min(m,min(niz1[i],niz2[i]));
		}
		for(int i=0;i<niz1.size();i++)
		{
			niz1[i]-=m;
			niz2[i]-=m;
		}
	}
	if(niz1[0]>niz2[0])
	cout<<"No";
	else
	cout<<"Yes";
    return 0;
}