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<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
题意是01背包问题，给定一堆硬币，和一个目标值，选出和为目标值的最小序列。
数据量还是比较小的，直接暴力深搜了，最后一个点的情况是所有的硬币加起来都没有给定的目标值大，所以特判一下就能过了。
主要是目标值给的太小了，如果稍微大一点深搜就很吃力了。
#include<bits/stdc++.h>
using namespace std;
void dfs(vector<int>& coins, int m, int cur, vector<int>& path, vector<int>& ret){
if (m < 0||ret.size()) return;
if (m == 0){
ret = path;
return;
}
for (int i = cur + 1; i < coins.size(); i++){
path.push_back(coins[i]);
dfs(coins, m - coins[i], i, path, ret);
path.pop_back();
}
}
int main()
{
int n, m;
cin >> n >> m;
vector<int>coins(n);
for (int i = 0; i < n; i++)
cin >> coins[i];
if(accumulate(coins.begin(),coins.end(),0)<m){
cout << "No Solution";
return 0;
}
sort(coins.begin(), coins.end());
vector<int>path, ret;
dfs(coins, m, -1, path, ret);
if (ret.size() == 0)
cout << "No Solution";
else
for (int i = 0; i <ret.size(); i++)
cout <<( i==0 ? "" : " ") << ret[i];
return 0;
}