Cut Ribbon
Polycarpus has a ribbon, its length is n. He wants to cut the ribbon in a way that fulfils the following two conditions:
- After the cutting each ribbon piece should have length a, b or c.
- After the cutting the number of ribbon pieces should be maximum.
Help Polycarpus and find the number of ribbon pieces after the required cutting.
The first line contains four space-separated integers n, a, b and c (1 ≤ n, a, b, c ≤ 4000) — the length of the original ribbon and the acceptable lengths of the ribbon pieces after the cutting, correspondingly. The numbers a, b and c can coincide.
Print a single number — the maximum possible number of ribbon pieces. It is guaranteed that at least one correct ribbon cutting exists.
5 5 3 2
2
7 5 5 2
2
In the first example Polycarpus can cut the ribbon in such way: the first piece has length 2, the second piece has length 3.
In the second example Polycarpus can cut the ribbon in such way: the first piece has length 5, the second piece has length 2.
#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <unordered_set>
#include <unordered_map>
#define ll long long
#define mod 998244353
using namespace std;
int dir[][] = { {,},{,-},{-,},{,} }; int main()
{
int n;
cin >> n;
vector<int> a(),dp(n+);
for (int i = ; i < ; i++)
cin >> a[i];
sort(a.begin(), a.end());
if (a[] <= n) dp[a[]] = ;
if (a[] <= n) dp[a[]] = ;
if (a[] <= n) dp[a[]] = ;
for (int i = ; i <= n; i++)
{
int ans = dp[i];
if (i > a[] && dp[i - a[]] != ) ans = max(ans, dp[i - a[]] + );
if (i > a[] && dp[i - a[]] != ) ans = max(ans, dp[i - a[]] + );
if (i > a[] && dp[i - a[]] != ) ans = max(ans, dp[i - a[]] + );
dp[i] = ans;
}
cout << dp[n] << endl;
return ;
}
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