百练 宠物小精灵之收服

思路：

dp。

实现：

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;

int N, M, K;
int a[105], b[105], dp[1005][505];

void solve()
{
    for (int i = K - 1; i >= 0; i--)
    {
        for (int j = N; j >= 0; j--)
        {
            for (int k = M; k >= 0; k--)
            {
                dp[j][k] = dp[j][k];
                if (j >= a[i] && k >= b[i])
                {
                    dp[j][k] = max(dp[j][k], dp[j - a[i]][k - b[i]] + 1);
                }
            }
        }
    }
    int maxN = 0, costM = 0;
    for (int i = 0; i < M; i++)
    {
        if (dp[N][i] > maxN)
        {
            maxN = dp[N][i];
            costM = i;
        }
    }
    cout << maxN << " " << M - costM << endl;
}
int main()
{
    cin >> N >> M >> K;
    for (int i = 0; i < K; i++)
    {
        cin >> a[i] >> b[i];
    }
    solve();
    return 0;
}