6.2 PAT 甲级 1068 Find More Coins

Eva loves to collect coins from all over the universe, including some other planets like Mars. One day she visited a universal shopping mall which could accept all kinds of coins as payments. However, there was a special requirement of the payment: for each bill, she must pay the exact amount. Since she has as many as 10​4​​ coins with her, she definitely needs your help. You are supposed to tell her, for any given amount of money, whether or not she can find some coins to pay for it.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive numbers: N (≤10​4​​, the total number of coins) and M (≤10​2​​, the amount of money Eva has to pay). The second line contains N face values of the coins, which are all positive numbers. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the face values V​1​​≤V​2​​≤⋯≤V​k​​ such that V​1​​+V​2​​+⋯+V​k​​=M. All the numbers must be separated by a space, and there must be no extra space at the end of the line. If such a solution is not unique, output the smallest sequence. If there is no solution, output "No Solution" instead.

Note: sequence {A[1], A[2], ...} is said to be "smaller" than sequence {B[1], B[2], ...} if there exists k≥1 such that A[i]=B[i] for all i

Sample Input 1:

8 9
5 9 8 7 2 3 4 1

Sample Output 1:

1 3 5

Sample Input 2:

4 8
7 2 4 3

Sample Output 2:

No Solution

是一题挺典型的动态规划。

我卡了挺久的点：之所以要将背包里的硬币从大到小排序，是因为需要的序列是按照字典序从小到大输出，即f[n-1][m-w[n]]这样从背包的最后一个开始减，所以越小的要放到背包的越后面。

#include
#include
using namespace std;

const int N = 10010,M = 110;

int n,m,w[N];
/*
f[i][j]表示选到第i个硬币,总面额为j的组成方式存不存在
可以分为不选第i个硬币：f[i][j] = f[i - 1][j]
选第i个硬币：f[i][j] = f[i - 1][j - w[i]]
*/
bool f[N][M];

int main()
{
    cin >> n >> m;
    
    for(int i = 1; i <= n; i ++ ) cin >> w[i];
    sort(w + 1, w + n + 1, greater());       //***
    
    f[0][0] = true;
    for(int i = 1; i <= n; i ++ )
        for(int j = 0; j <= m; j ++ )
        {
            f[i][j] = f[i - 1][j];
            if(j >= w[i])
                f[i][j] |= f[i - 1][j - w[i]];
        }
    
    if(!f[n][m])
        cout << "No Solution" << endl;
    else
    {
        bool is_first = true;
        while(n)
        {
            if(f[n - 1][m - w[n]] && m >= w[n] )
            {
                if(is_first) is_first = false;
                else
                    cout << ' ';
                cout << w[n];
                m -= w[n];
            }
            n --;
        }
    }
}