1068 Find More Coins (30point(s)) - C語言 PAT 甲級

1068 Find More Coins (30point(s))

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 (≤10⁴, the total number of coins) and M (≤10², 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 V₁≤V₂≤⋯≤V_k such that V₁≤V₂≤⋯≤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<k, and A[k] < B[k].

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

題目大意：

輸入 N 個硬幣，要求選擇若干硬幣恰好等於支付金額 M，選擇的硬幣越小越好

設計思路：

0-1 揹包問題生動講解
遞歸+枚舉的方法

• 需要輸出序列較小的答案，先對硬幣排序從大到小，
• 然後按照 0-1 揹包問題求解，即先選大面額的硬幣，後續遇到小面額符合條件的就選擇小面額
• 除了揹包問題所需的 dp[] 遞推數組，還需要額外建立 choice[][] 數組表示硬幣選擇情況

編譯器：C (gcc)

#include <stdio.h>

int cmp(const void *a, const void *b)
{
        return *((int *)b) - *((int *)a);
}

int main(void)
{
        int n, m;
        int dp[10010] = {0}, v[10010] = {0}, select[10010][110] = {0};
        int i, j;

        scanf("%d%d", &n, &m);
        for (i = 1; i <= n; i++)
                scanf("%d", &v[i]);
        qsort(v + 1, n, sizeof(v[0]), cmp);
        for (i = 1; i <= n; i++) {
                for (j = m; j >= v[i]; j--) {
                        if (dp[j] <= dp[j - v[i]] + v[i]) {
                                select[i][j] = 1;
                                dp[j] = dp[j - v[i]] + v[i];
                        }
                }
        }
        if (dp[m] != m) {
                printf("No Solution");
        } else {
                int f = 0;
                i = n;
                j = m;
                while (j > 0) {
                        if (select[i][j] == 1) {
                                printf("%s%d", f ? " " : "", v[i]);
                                f = 1;
                                j -= v[i];
                        }
                        i--;
                }
        }
        return 0;
}