牛客小白月赛7 G - CSL分苹果（dp）

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思路：dp[j]来表示当最多可容纳 j 质量的质量，则dp方程：

#include<bits/stdc++.h>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define LL long long
int main()
{
    int n, sum, w[110], dp[10010];
    while (cin >> n)
    {
        sum = 0;
        memset(dp, 0, sizeof(dp));
        for (int i = 0; i < n; i++)
        {
            cin >> w[i];
            sum += w[i];
        }
        for (int i = 0; i < n; i++)
            for(int j = sum / 2; j >= w[i]; j--)
                dp[j] = max(dp[j], dp[j - w[i]] + w[i]);
        cout << dp[sum / 2] << " " << sum - dp[sum / 2] << endl;
    }
    return 0;
}