[pat]1068 Find More Coins

满背包问题，把体积和价值看成相等的。用滚动数组优化，然后额外开辟一个choice数组来记录每次的选择，然后回溯打印。因为要按字典序，先把价值进行排序。假如选最小的商品能装满m的话，那就把判断条件改成大于等于，然后最后来

#include<bits/stdc++.h>
using namespace std;
int dp[105];
int w[10005];
bool flag[10005];
bool choice[10005][105];
int n, m;
bool cmp(int a, int b)
{
    return a > b;
}
int main()
{
    memset(flag, false, sizeof(flag));
    fill(choice[0], choice[0] +105, false);
    fill(dp, dp +105, 0);
    scanf("%d %d", &n, &m);
    int i, j;
    for (i = 1; i <= n; i++)
    {
        scanf("%d", &w[i]);
    }
    sort(w+1, w + n+1,cmp);
    for (i = 1; i <= n; i++)
    {
        for (j = m; j >= w[i]; j--)
        {
            if (dp[j-w[i]]+w[i]>=dp[j])
            {
                dp[j] = dp[j - w[i]] + w[i];
                choice[i][j] = true;
            }
        }
    }
    int x=n, y=m;
    int num = 0;
    if (dp[m] != m)
    {
        printf("No Solution\n");
    }
    else
    {
        while (x >= 1)
        {
            if (choice[x][y] == true)
            {
                flag[x] = true;//下标
                num++;
                y -= w[x];
            }
            x--;
        }
        for (i = n; i >= 1; i--)
        {
            if (flag[i] == true)
            {
                if (num - 1 > 0)
                {
                    printf("%d ", w[i]);
                    num--;
                }
                else
                    printf("%d\n", w[i]);
            }
        }
    }
}

选择最小的那个。