189A Cut Ribbon(dp) - 码农教程

题意

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

一道完全背包的题目，由于之前只是接触了01背包，没有向下扩展下去，所以没有看出来。状态转移方程是dp[j]=max(dp[j],dp[j-a[i]]+1)。需要注意的是，要把dp[1~n]初始化为-INF，否则可能会出现没装满的情况。

AC代码

#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;
}