轉化一下就是01背包 CodeForces 433A - 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,?...,?wn (wi?=?100 or 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

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.

題目就是說一個人買了一大堆的apple，然後給出每個蘋果的質量，要求將這些蘋果分成等質量的兩堆，不就是動態規劃嗎？？？！！！ 講一下思路：先將質量全部加起來為sum，如果sum余2不為零直接NO，然後sum/=2；接下來就是01背包了，只要dp[sum/2]==sum/2；就行了 嗯，貼下代碼：

#include
#include
#include
#include
using namespace std;

int main()
{
    int i,j,k;
    int t,n,m;
    __int64 sum;
    int liu[106],dp[250000];
    while(scanf("%d",&n)!=EOF)
    {
        sum=0;
        memset(liu,0,sizeof(liu));
        memset(dp,0,sizeof(dp));
        for(i=1;i<=n;i++)
        {
            scanf("%d",&liu[i]);
            sum+=liu[i];
        }
        if(sum%2!=0)
        {
            printf("NO\n");
            continue;
        }
        sum/=2;
        //printf("%d\n",sum);
        for(i=1;i<=n;i++)
        {
            for(j=sum;j>=liu[i];j--)
                dp[j]=max(dp[j],dp[j-liu[i]]+liu[i]);
        }
        if(dp[sum]==sum)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}

發現bug就講吧，謝謝；