1068 Find More Coins （30 分)

2019-04-02

1068 Find More Coins （30 分)

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 104 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:

1 2	8 9 5 9 8 7 2 3 4 1

Sample Output 1:

1 3 5

Sample Input 2:

1 2	4 8 7 2 4 3

Sample Output 2:

1	No Solution

作者: CHEN, Yue

单位: 浙江大学

时间限制: 200 ms

内存限制: 64 MB

代码长度限制: 16 KB

题目大意

从一堆硬币中，选出和加起来为m的硬币。并且硬币的序列要最小。如果没有相应的方案，输出”No Solution”。

分析

采用深搜，加上一些优化。首先对输入进行排序，硬币由小选到大，这样找到的第一个解就是最小的。第二记录所有硬币的和，如果小于m，那么直接输出”No Solution”。

代码

#include <iostream>
#include <vector>
#include <algorithm>
#include <deque>
#include <map>
#include <set>
#include <cstring>
#include <cmath>
#include <sstream>

using namespace std;
int n, m;

vector<int> temp;

int a[10005];

void dfs(int i, int total) {
    if (i == n)
        return;
    if (total + a[i] == m) {
        temp.push_back(a[i]);
        cout << temp[0];
        for (int i = 1; i < temp.size(); i++)
            cout << " " << temp[i];
        exit(0);
    }
    for (int j = 0; j < 2; j++) {
        if (j == 0) {//选此硬币
            temp.push_back(a[i]);
            dfs(i + 1, total + a[i]);
            temp.pop_back();
        } else if (total + a[i] < m)//不选此硬币
            dfs(i + 1, total);

    }
}

int main() {
    int sum = 0;
    cin >> n >> m;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
        sum += a[i];
    }
    if (sum < m)
        cout << "No Solution" << endl;
    else {
        sort(a, a + n);
        dfs(0, 0);
        cout << "No Solution" << endl;
    }
    return 0;
}

五、其他

附上一个同样是深搜，写的很好的写法：https://blog.csdn.net/sinat_29278271/article/details/48022061
这道题还有动态规划的解法，参考：
1.https://blog.csdn.net/xyt8023y/article/details/47255241
2.https://blog.csdn.net/tiantangrenjian/article/details/17334201
3.https://blog.csdn.net/gzxcyy/article/details/14025949