189A Cut Ribbon(dp)

題意

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.

Input
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.

Output
Print a single number — the maximum possible number of ribbon pieces. It is guaranteed that at least one correct ribbon cutting exists.

Examples
input
5 5 3 2
output
2
input
7 5 5 2
output
2

思路

一道完全揹包的題目，由於之前只是接觸了01揹包，沒有向下擴展下去，所以沒有看出來。狀態轉移方程是dp[j]=max(dp[j],dp[j-a[i]]+1)。需要注意的是，要把dp[1~n]初始化爲-INF，否則可能會出現沒裝滿的情況。

AC代碼

#include<bits/stdc++.h>
#define x first
#define y second
#pragma GCC optimize(2)
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC optimize(3)
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC target("sse3","sse2","sse")
#pragma GCC target("avx","sse4","sse4.1","sse4.2","ssse3")
#pragma GCC target("f16c")
#pragma GCC optimize("inline","fast-math","unroll-loops","no-stack-protector")
#pragma GCC diagnostic error "-fwhole-program"
#pragma GCC diagnostic error "-fcse-skip-blocks"
#pragma GCC diagnostic error "-funsafe-loop-optimizations"
#pragma GCC diagnostic error "-std=c++14"
#define IOS ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
typedef unsigned long long ULL;
typedef pair<int,int> PII;
typedef long long LL;
const int N=5005;
const int INF=0x3f3f3f3f;
const int MOD=998244353;
int a[N];
int dp[N];
int main(){
    IOS;
    int a[4];
    int n;cin>>n;
    for(int i=1;i<=3;i++) cin>>a[i];
    for(int i=1;i<=n;i++) dp[i]=-INF;
    dp[0]=0;
    for(int i=1;i<=3;i++){
        for(int j=a[i];j<=n;j++){
            dp[j]=max(dp[j],dp[j-a[i]]+1);
        }
    }
    cout<<dp[n]<<endl;
    return 0;
}