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  • Codeforces Round #119 (Div. 2)A. Cut Ribbon

    A. Cut Ribbon
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    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 ab 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 nab 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 ab 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
    Note

    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.

    这题n^2枚举可以解决,但是也可以用dp O(n)啊

    dp得到的数。

    dp[i]代表 组成i 需要的最多ribbon块数。

    /* ***********************************************
    Author        :guanjun
    Created Time  :2016/10/7 18:22:02
    File Name     :cf119a.cpp
    ************************************************ */
    #include <bits/stdc++.h>
    #define ull unsigned long long
    #define ll long long
    #define mod 90001
    #define INF 0x3f3f3f3f
    #define maxn 10010
    #define cle(a) memset(a,0,sizeof(a))
    const ull inf = 1LL << 61;
    const double eps=1e-5;
    using namespace std;
    priority_queue<int,vector<int>,greater<int> >pq;
    struct Node{
        int x,y;
    };
    struct cmp{
        bool operator()(Node a,Node b){
            if(a.x==b.x) return a.y> b.y;
            return a.x>b.x;
        }
    };
    
    bool cmp(int a,int b){
        return a>b;
    }
    int dp[4010];
    int main()
    {
        #ifndef ONLINE_JUDGE
        //freopen("in.txt","r",stdin);
        #endif
        //freopen("out.txt","w",stdout);
        int n,a,b,c;
        while(cin>>n>>a>>b>>c){
            cle(dp);
            dp[a]=dp[b]=dp[c]=1;
            for(int i=1;i<=n;i++){
                if(i>a&&dp[i-a])dp[i]=max(dp[i],dp[i-a]+1);
                if(i>b&&dp[i-b])dp[i]=max(dp[i],dp[i-b]+1);
                if(i>c&&dp[i-c])dp[i]=max(dp[i],dp[i-c]+1);
            }
            cout<<dp[n]<<endl;
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/pk28/p/5936486.html
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