codeforces --- Round #248 (Div. 2) A. Kitahara Haruki's Gift

Kitahara Haruki has bought n apples for Touma Kazusa and Ogiso Setsuna. Now he wants to divide all the apples between the friends.

Each apple weights 100 grams or 200 grams. Of course Kitahara Haruki doesn't want to offend any of his friend. Therefore the total weight of the apples given to Touma Kazusa must be equal to the total weight of the apples given to Ogiso Setsuna.

But unfortunately Kitahara Haruki doesn't have a knife right now, so he cannot split any apple into some parts. Please, tell him: is it possible to divide all the apples in a fair way between his friends?

Input

The first line contains an integer n (1 ≤ n ≤ 100) — the number of apples. The second line contains n integers w₁, w₂, ..., w_n (w_i = 100 or w_i = 200), where w_i is the weight of the i-th apple.

Output

In a single line print "YES" (without the quotes) if it is possible to divide all the apples between his friends. Otherwise print "NO" (without the quotes).

Sample test(s)

Input

3
100 200 100

Output

YES

Input

4
100 100 100 200

Output

NO

Note

In the first test sample Kitahara Haruki can give the first and the last apple to Ogiso Setsuna and the middle apple to Touma Kazusa.

【题目大意】

有两种面值的钞票，100元或200元的各有若干张，现在要将其分给两个人，要你判断能否均分。

【题目分析】

既然判断能否均分，那就来做分类讨论：

伪代码如下：

if(200元的能均分)

{

if(100元的也能均分)

输出YES;

else 输出NO;

}

else

{

if(100元的数量<2)

输出NO;

else

{

100元的数量-=2；//(为了将200元的凑成2的倍数，当然也可不减)

if(100元的数量能均分)

输出YES;

else 输出NO；

}

结束；

AC代码1:

#include<iostream>
using namespace std;
int main()
{
    int n;
    cin>>n;
    int i,a;
    int cnt1=0,cnt2=0;
    for(i=0;i<n;i++)
    {
        cin>>a;
        if(a==100)
            cnt1++;
        else cnt2++;
    }
    if(cnt2%2==0)
    {
        if(cnt1%2==0)
            cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
    }
    else
    {
        if(cnt1<2)
            cout<<"NO"<<endl;
        else
        {
            cnt1-=2;
            if(cnt1%2==0)
               cout<<"YES"<<endl;
            else cout<<"NO"<<endl;
        }
    }
    return 0;
}

ac代码2：

#include<iostream>
using namespace std;
int main()
{
    int n;
    cin>>n;
    int i,a;
    int cnt1=0,cnt2=0;
    for(i=0;i<n;i++)
    {
        cin>>a;
        if(a==100)
            cnt1++;
        else cnt2++;
    }
    if(cnt2%2==0)
    {
        if(cnt1%2==0)
            cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
    }
    else
    {
        if(cnt1<2)
            cout<<"NO"<<endl;
        else
        {
            if(cnt1%2==0)
               cout<<"YES"<<endl;
            else cout<<"NO"<<endl;
        }
    }
    return 0;
}

ac代码3：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
using namespace std;

long n,a,h,t,i;

int main()
{
    while(~scanf("%ld",&n))
    {
        h=t=0;
        for(i=0;i<n;i++)
        {
            scanf("%ld",&a);
            if(a==100)
                h++;
            else
                t++;
        }
        t=t%2;
        h-=t*2;
        if(h<0 || h%2==1)
            printf("NO
");
        else
            printf("YES
");
    }
    return 0;
}

ac代码4:

#include <iostream>
#include <stdio.h>

using namespace std;

int main()
{
    int t;
    cin >> t;
    int a=0;
    int b=0;
    int e;
    for (int i=0;i<t;i++)
    {
        cin >> e;
        if (e==200)a++;
        else b++;
    }
    int c=0,d=0;
    while (a>0)
    {
        if (c<d)
        {
            c+=200;
        }
        else
        {
            d+=200;
        }
        a--;
    }
    while(b>0)
    {
        if (c<d)
        {
            c+=100;
        }
        else
        {
            d+=100;
        }
        b--;
    }
    if (c==d)cout << "YES" << endl;
    else cout << "NO" << endl;
    return 0;
}