zoukankan      html  css  js  c++  java
  • Fruit Feast(暴力)(动态规划)

    Fruit Feast

    时间限制: 1 Sec  内存限制: 64 MB
    提交: 64  解决: 18
    [提交][状态][讨论版]

    题目描述

    Bessie has broken into Farmer John's house again! She has discovered a pile of lemons and a pile of oranges in the kitchen (effectively an unlimited number of each), and she is determined to eat as much as possible.

    Bessie has a maximum fullness of T(1≤T≤5,000,000). Eating an orange increases her fullness by A, and eating a lemon increases her fullness by B (1≤A,B≤T). Additionally, if she wants, Bessie can drink water at most one time, which will instantly decrease her fullness by half (and will round down).

    Help Bessie determine the maximum fullness she can achieve!

    输入

    The first (and only) line has three integers T, A, and B.

    输出

     A single integer, representing the maximum fullness Bessie can achieve.

    样例输入

    8 5 6
    

    样例输出

    8
    【分析】在此提供两种做法,一种暴力,一种动态规划。
    #include <iostream>
    #include <cstdio>
    #include <cstdlib>
    #include <cmath>
    #include <algorithm>
    #include <climits>
    #include <cstring>
    #include <string>
    #include <set>
    #include <map>
    #include <queue>
    #include <stack>
    #include <vector>
    #include <list>
    #include<functional>
    #define mod 1000000007
    #define inf 0x3f3f3f3f
    #define pi acos(-1.0)
    using namespace std;
    typedef long long ll;
    const int N=5000005;
    const int M=15005;
    ll sum=1;
    int n,m;
    int vis[N];
    int maxn=0;
    struct man
    {
      int num,use;
    };
    int main()
    {
        int a,b;
        memset(vis,0,sizeof(vis));
        scanf("%d%d%d",&n,&a,&b);
        queue<man>q;
        man A;A.num=a;A.use=0;
        man B;B.num=b;B.use=0;
        q.push(A);q.push(B);
        vis[a]=1;vis[b]=1;
        while(!q.empty()){
            man t=q.front();
            q.pop();
            if(t.num+a>n)maxn=max(maxn,t.num);
            if(t.num+a<=n&&vis[t.num+a]==0){
                vis[t.num+a]=1;
                man k;k.num=t.num+a;k.use=t.use;
                q.push(k);
            }
            if(t.num+b>n)maxn=max(maxn,t.num);
            if(t.num+b<=n&&vis[t.num+b]==0){
                vis[t.num+b]=1;
                    man k;k.num=t.num+b;k.use=t.use;
                    q.push(k);
            }
            if(t.use==0&&vis[t.num/2]==0){
                vis[t.num/2]=1;
            man k;k.num=t.num/2;k.use=1;
                q.push(k);
            }
        }
        printf("%d
    ",maxn);
        return 0;
    }
    暴力
    #include <iostream>
    #include <cstdio>
    #include <cstdlib>
    #include <cmath>
    #include <algorithm>
    #include <climits>
    #include <cstring>
    #include <string>
    #include <set>
    #include <map>
    #include <queue>
    #include <stack>
    #include <vector>
    #include <list>
    #include<functional>
    #define mod 1000000007
    #define inf 0x3f3f3f3f
    #define pi acos(-1.0)
    using namespace std;
    typedef long long ll;
    const int N=5000005;
    const int M=15005;
    ll sum=1;
    int n,m;
    int vis[N];
    int maxn=0;
    int dp[N];
    int main()
    {
        int a,b;
        scanf("%d%d%d",&n,&a,&b);
        dp[0]=1;
        for(int i=a;i<=n;i++)dp[i]|=dp[i-a];
        for(int i=b;i<=n;i++)dp[i]|=dp[i-b];
        for(int i=0;i<=n;i++)dp[i/2]|=dp[i];
        for(int i=a;i<=n;i++)dp[i]|=dp[i-a];
        for(int i=b;i<=n;i++)dp[i]|=dp[i-b];
        for(a =n;dp[a]==0;a--);
        cout<<a<<endl;
        return 0;
    }
    动态规划
  • 相关阅读:
    BZOJ2527[Poi2011]Meteors——整体二分+树状数组
    [UOJ422][集训队作业2018]小Z的礼物——轮廓线DP+min-max容斥
    BZOJ4817[Sdoi2017]树点涂色——LCT+线段树
    BZOJ4269再见Xor——高斯消元解线性基
    BZOJ4241历史研究——回滚莫队
    [十二省联考2019]字符串问题——后缀自动机+parent树优化建图+拓扑序DP+倍增
    [十二省联考2019]异或粽子——可持久化trie树+堆
    [CF594E]Cutting the Line
    [CF1246F]Cursor Distance
    [CF1246E]To Make 1
  • 原文地址:https://www.cnblogs.com/jianrenfang/p/5760362.html
Copyright © 2011-2022 走看看