CF#248DIV2:A. Kitahara Haruki's Gift_html/css_WEB-ITnose

WBOY
Release: 2016-06-24 12:03:56
Original
1146 people have browsed it

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,?...,?wn (wi?=?100or wi?=?200), where wi 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

3100 200 100
Copy after login

output

YES
Copy after login

input

4100 100 100 200
Copy after login

output

NO
Copy after login

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.


好久没做CF了,这次又是打了个酱油

#include <stdio.h>#include <string.h>#include <math.h>#include <algorithm>using namespace std;int a[105];int main(){    int n,i,sum1,sum2,sum;    while(~scanf("%d",&n))    {        sum1 = sum2 = sum = 0;        for(i = 0; i<n; i++)        {            scanf("%d",&a[i]);            sum+=a[i];            if(a[i] == 100)                sum1++;            else                sum2++;        }        if(n == 1)        {            printf("NO\n");            continue;        }        sort(a,a+n);        sum/=2;        if(sum%100!=0 || sum1%2)            printf("NO\n");        else        {            if(sum2%2 && sum1==0)                printf("NO\n");            else                printf("YES\n");        }    }    return 0;}
Copy after login


Related labels:
source:php.cn
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template
About us Disclaimer Sitemap
php.cn:Public welfare online PHP training,Help PHP learners grow quickly!