Buying Apples!

本文介绍了一个基于完全背包问题的应用案例,旨在帮助用户计算购买指定数量苹果所需的最低金额,同时考虑到商品包装规格及库存情况。

Buying Apples!

 

Harish went to a supermarket to buy exactly ‘k’ kilograms apples for his ‘n’ friends. The supermarket was really weird. The pricing of items was very different. He went to the Apples section and enquired about the prices. The salesman gave him a card in which he found that the prices of apples were not per kg. The apples were packed into covers, each containing ‘x’ kg of apples, x > 0 and ‘x’ is an integer. An ‘x’ kg packet would be valued at ‘y’ rupees. So, the placard contained a table with an entry ‘y’ denoting the price of an ‘x’ kg packet. If ‘y’ is -1 it means that the corresponding packet is not available. Now as apples are available only in packets, he decides to buy atmost ‘n’ packets for his ‘n’ friends i.e he will not buy more than n packets of apples.

Harish likes his friends a lot and so he does not want to disappoint his friends. So now, he will tell you how many friends he has and you have to tell him the minimum amount of money he has to spend for his friends.

Input

The first line of input will contain the number of test cases, C.

Each test case will contain two lines.

The first line containing N and K, the number of friends he has and the amount of Apples in kilograms which he should buy.

The second line contains K space separated integers in which the ith integer specifies the price of a ‘i’kg apple packet. A value of -1 denotes that the corresponding packet is unavailable.

  • 0 < N <= 100
  • 0 < K <= 100
  • 0 < price <= 1000

Output

The output for each test case should be a single line containing the minimum amount of money he has to spend for his friends. Print -1 if it is not possible for him to satisfy his friends.

Sample I/O

Input:
2
3 5
-1 -1 4 5 -1
5 5
1 2 3 4 5

Output:
-1
5

Explanation of test cases:

1) As there are only 3 and 4kg packets in the shop, he will not be able to satisfy his friends as he would not be able to buy exactly 5kg of apples.

2) He can buy five 1kg packet as he has to buy 5 kg. So the min money he should spend is 5.

【分析】完全背包

有N个朋友,要买k千克苹果,超市里有1~k千克的苹果,各a[i]元,问能否买到k千克的苹果,所花的钱最少,并且不超过N件(然而N没用。。。)。



#include <iostream>
#include <cstdio>#include <cmath>#include <cstring>using namespace std;#define cl(a,b) memset(a,b,sizeof a);const int maxn = 5051;const int INF = 0x3f3f3f3f;int a[101];int dp[maxn];int num[maxn];int main(){ int t; scanf("%d",&t); while(t--){ int n,m; for(int i=0;i<maxn;i++) dp[i] = INF; cl(num,0); scanf("%d%d",&n,&m); for(int i=1;i<=m;i++){ scanf("%d",a+i); if(a[i]!=-1){ dp[i] = a[i]; num[i] = 1; } } for(int i=1;i<=m;i++){ if(a[i]!=-1) for(int j=0;j<m;j++){ if(dp[j]!=INF){ if(dp[i+j] > dp[j]+a[i]&&num[j]+1<=n){ dp[i+j] = dp[j] + a[i]; num[i+j] = num[j] + 1; } } } } if(dp[m]==INF) dp[m] = -1; printf("%d\n",dp[m]); } return 0;}


MATLAB代码实现了一个基于多种智能优化算法优化RBF神经网络的回归预测模型,其核心是通过智能优化算法自动寻找最优的RBF扩展参数(spread),以提升预测精度。 1.主要功能 多算法优化RBF网络:使用多种智能优化算法优化RBF神经网络的核心参数spread。 回归预测:对输入特征进行回归预测,适用于连续值输出问题。 性能对比:对比不同优化算法在训练集和测试集上的预测性能,绘制适应度曲线、预测对比图、误差指标柱状图等。 2.算法步骤 数据准备:导入数据,随机打乱,划分训练集和测试集(默认7:3)。 数据归一化:使用mapminmax将输入和输出归一化到[0,1]区间。 标准RBF建模:使用固定spread=100建立基准RBF模型。 智能优化循环: 调用优化算法(从指定文件夹中读取算法文件)优化spread参数。 使用优化后的spread重新训练RBF网络。 评估预测结果,保存性能指标。 结果可视化: 绘制适应度曲线、训练集/测试集预测对比图。 绘制误差指标(MAE、RMSE、MAPE、MBE)柱状图。 十种智能优化算法分别是: GWO:灰狼算法 HBA:蜜獾算法 IAO:改进天鹰优化算法,改进①:Tent混沌映射种群初始化,改进②:自适应权重 MFO:飞蛾扑火算法 MPA:海洋捕食者算法 NGO:北方苍鹰算法 OOA:鱼鹰优化算法 RTH:红尾鹰算法 WOA:鲸鱼算法 ZOA:斑马算法
评论
成就一亿技术人!
拼手气红包6.0元
还能输入1000个字符
 
红包 添加红包
表情包 插入表情
 条评论被折叠 查看
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值