hetao8554411 LV 7 @ 2023-7-12 14:23:55

`金币Lv.1`

1.课听了吧?

听了，这道题是作业
没听

直接把作业拷过来就行

#include <iostream>
#include <algorithm>
#include <iomanip>

using namespace std;

struct Coin
{
    double m,v,avg;
} a[10005];

bool cmp(Coin a,Coin b)
{
    return a.avg > b.avg;
}

int main()
{
    int n, t;
    cin >> n >> t;
    for (int i = 0; i < n; i++)
    {
        cin >> a[i].m >> a[i].v;
        a[i].avg = a[i].v/a[i].m;
    }
  
    sort(a, a + n, cmp);
  
    int pos = 0;
    double sum = 0;
    while (t > 0)
    {
        double temp;
  
        if(t > a[pos].m)
            temp = a[pos].m;
        else
            temp = t;
    
        sum += temp * a[pos].avg,t -= temp;
  
        pos++;
    }
  
    cout << fixed << setprecision(2) << sum;
  
    return 0;
}

hetao8554411 20230712

hetao2934713 @ 2023-7-27 18:57:17

巧了，我喜欢在做完作业后看一眼下节课的题，正好看见作业是有一句“分割完后价值不变”就想到了这一道题，所以直接开浏览器，~~复制了题解~~
hetao2587951 @ 2024-4-28 11:41:01

@我也是

7

hetao8592334 LV 9 @ 2023-12-9 18:59:02

#include <iostream>
#include <algorithm>
#include <iomanip>

using namespace std;

struct Coin
{
    double m,v,avg;
} a[10005];

bool cmp(Coin a,Coin b)
{
    return a.avg > b.avg;
}

int main()
{
    int n, t;
    cin >> n >> t;
    for (int i = 0; i < n; i++)
    {
        cin >> a[i].m >> a[i].v;
        a[i].avg = a[i].v/a[i].m;
    }
  
    sort(a, a + n, cmp);
  
    int pos = 0;
    double sum = 0;
    while (t > 0)
    {
        double temp;
  
        if(t > a[pos].m)
            temp = a[pos].m;
        else
            temp = t;
    
        sum += temp * a[pos].avg,t -= temp;
  
        pos++;
    }
  
    cout << fixed << setprecision(2) << sum;
  
    return 0;
}

没上课：好难的部分背包上课了：简单的部分背包

5

hetao1357085 LV 8 @ 2023-9-10 19:26:31

#include <iostream>//hetao1357085
#include <algorithm>
#include <iomanip>

using namespace std;

struct Coin
{
    double m,v,avg;
} a[10005];

bool cmp(Coin a,Coin b)
{
    return a.avg > b.avg;
}

int main()
{
    int n, t;
    cin >> n >> t;
    for (int i = 0; i < n; i++)
    {
        cin >> a[i].m >> a[i].v;
        a[i].avg = a[i].v/a[i].m;
    }
  
    sort(a, a + n, cmp);
  
    int pos = 0;
    double sum = 0;
    while (t > 0)
    {
        double temp;
  
        if(t > a[pos].m)
            temp = a[pos].m;
        else
            temp = t;
    
        sum += temp * a[pos].avg,t -= temp;
  
        pos++;
    }
  
    cout << fixed << setprecision(2) << sum;
  
    return 0;
}

Jntm,给个赞吧

4

hetao1889320 LV 6 @ 2023-12-9 13:25:11

e.....这道题怎么说呢，说是难题吧你上了课就秒变签到题，说是签到题吧，你不上课就是挺难的部分背包><~~```~~

#include <bits/stdc++.h>
using namespace std;
struct Crystal
{
double volume, power;
double avg;
};
Crystal a[10005];
bool cmp(Crystal a, Crystal b)
{
return a.avg > b.avg;
}
int main()
{
int n, v;
cin >> n >> v;
for (int i = 0; i < n; i++)
{
cin >> a[i].volume >> a[i].power;
a[i].avg = a[i].power/a[i].volume;
}
sort(a,a+n,cmp   );
int pos = 0;
double sum = 0;
while (v > 0)
{
double temp;
if(v > a[pos].volume)
temp = a[pos].volume;
else
temp = v;
sum+=temp*a[pos].avg;
v-=temp;
pos++;
}
cout << fixed << setprecision(2) << sum << endl;//保留两位小数输出，setprecision（2）的“2”是可以更改的
return 0;
}

2

TomAnderson LV 3 SU @ 2023-5-11 13:51:24

题目大意

存在n堆金币，第i堆金币的总重量和总价值是mi, vi。存在承重为T的背包，金币可切割，问最多可以拿走多少价值的金币。

完整思路

将每堆金币的单位价值求出来，并按照单位价值从大到小排序，一个个加入到背包中。

核心代码


//对单位价值排序
struct node
{             // 定义结构体
    double w; // 重量
    double v; // 价值
    double p; // 性价比
};
bool cmp(node a, node b)
{
    return a.p > b.p; // 性价比从大到小排序
}


//加入到背包中
int pos = 0;
double sum = 0;
while (c > 0)
{
    double temp = min(a[pos].w, c);
    c -= temp;
    sum += temp * a[pos].p;
    pos++;
}

1

hetao7922605 LV 4 @ 2024-4-27 19:17:59

cpp

#include <iostream> #include <algorithm> #include <iomanip>

using namespace std;

struct Coin { double m,v,avg; } a[10005];

bool cmp(Coin a,Coin b) { return a.avg > b.avg; }

int main() { int n, t; cin >> n >> t; for (int i = 0; i < n; i++) { cin >> a[i].m >> a[i].v; a[i].avg = a[i].v/a[i].m; }

sort(a, a + n, cmp);

int pos = 0;
double sum = 0;
while (t > 0)
{
    double temp;

    if(t > a[pos].m)
        temp = a[pos].m;
    else
        temp = t;

    sum += temp * a[pos].avg,t -= temp;

    pos++;
}

cout << fixed << setprecision(2) << sum;

return 0;

}

直接用上课的作业，就可以啦！别忘了用#include <iomanip>就好啦。思路不用说了吧，上过课的都知道。可供直接拷贝。

1

hetao13267928 LV 4 @ 2024-3-31 14:12:39

👍 
#include<iostream>
#include<algorithm>
#include <iomanip>
using namespace std;
int n,t;
struct Jinbi
{
    int m,v;
    double avg;
}j[105];
bool cmp(Jinbi x,Jinbi y)
{
    return x.avg>y.avg;
}
int main()
{
    cin>>n>>t;
    for(int i=1;i<=n;i++)
    {
        cin>>j[i].m>>j[i].v;
        j[i].avg=j[i].v*1.00/j[i].m;
    }
    sort(j+n,j+n+1,cmp);
    int pos=0;
    double sum=0;
    while(t>0)
    {
        int ans=min(t,j[pos].m);
        t-=ans;
        sum+=ans*j[pos].avg;
        pos++;
    }
    cout<<fixed<<setprecision(2)<<sum;
    return 0;
}

1

hetao16625820 LV 7 @ 2024-2-23 21:30:22

本题就是作业直接拷贝过来就行了

#include <iostream>
#include <algorithm>
using namespace std;
struct gold {
    int num, power;
    double avg;
};
gold a[10005];
bool cmp(gold a, gold b) {
    return a.avg > b.avg;
}
int main() {
    int n, m;
    cin >> n >> m;
    for (int i = 0; i < n; i++) {
        cin >> a[i].num >> a[i].power;
        a[i].avg = 1.0 * a[i].power / a[i].num;
    }
    sort(a, a + n, cmp);
    int pos = 0, temp;
    double sum = 0;
    while (m > 0) {
        temp = min(a[pos].num, m);
        if (m - temp > 0) {
            sum += a[pos].power;
        } else {
            sum += a[pos].avg * m;
        }
        m -= temp;
        pos++;
    }
    printf("%.2lf",sum);
    return 0;
}

1

hetao4976191 LV 6 @ 2023-10-4 13:06:43

#include <cstdio>
#include <algorithm>
using namespace std;
int n,pos;
double t,sum=0,tmp;
struct Gold{
    double m,v,avg;
}a[100];
bool cmp(Gold x,Gold y){
    return x.avg>y.avg;
}
int main(){
    scanf("%d%lf",&n,&t);
    for(int i=0;i<n;i++){
        scanf("%lf%lf",&a[i].m,&a[i].v);
        a[i].avg=a[i].v/a[i].m;
    }
    sort(a,a+n,cmp);
    while(t>0)tmp=min(t,a[pos].m),t-=tmp,sum+=tmp*a[pos++].avg;
    printf("%.2lf\n",sum);
    return 0;
}

1

hetao738467 LV 6 @ 2023-9-1 21:54:02

跟核桃上一样要“三省吾生”！

#include<bits/stdc++.h>
using namespace std;
int n,t;
struct golden
{
    int w;
    double v,sv;
}q[105];
    double sum;
bool cmp(golden x,golden y)
{ 
    return x.v>y.v;
}
int main()
{
    cin>>n>>t;
    for(int i=1;i<=n;i++)
    {
    cin>>q[i].w>>q[i].sv;
    q[i].v=q[i].sv/q[i].w;
    }
    sort(q+1,q+n+1,cmp);
    int pos=1;
    while(t>0)
    {
        int y=min(q[pos].w,t);
        t-=y;
        sum+=q[pos].v*y;
        pos++;
    }
    cout<<fixed<<setprecision(2)<<sum<<endl;
    return 0;
}

0

hetao7112278 LV 3 @ 2024-6-14 21:10:35

~~

#include <iostream>
#include <algorithm>
#include <iomanip>

using namespace std;

struct Coin
{
    double m,v,avg;
} a[10005];

bool cmp(Coin a,Coin b)
{
    return a.avg > b.avg;
}

int main()
{
    int n, t;
    cin >> n >> t;
    for (int i = 0; i < n; i++)
    {
        cin >> a[i].m >> a[i].v;
        a[i].avg = a[i].v/a[i].m;
    }
  
    sort(a, a + n, cmp);
  
    int pos = 0;
    double sum = 0;
    while (t > 0)
    {
        double temp;
  
        if(t > a[pos].m)
            temp = a[pos].m;
        else
            temp = t;
    
        sum += temp * a[pos].avg,t -= temp;
  
        pos++;
    }
  
    cout << fixed << setprecision(2) << sum;
  
    return 0;
}

```~~

0

hetao293087 LV 5 @ 2024-4-28 20:09:45

这道题.怎么说呢，直接就是~~作业换了个人名~~同类型的题好吧......

#include <bits/stdc++.h>
using namespace std;
int n;
double T;
struct coin
{
    double m,v,avg;
}a[105];
bool cmp(coin a,coin b)
{
    return a.avg > b.avg;
}
int main()
{
    cin >> n >> T;
    for(int i = 1;i <= n;i++)
    {
        cin >> a[i].m >> a[i].v;
        a[i].avg=a[i].v/a[i].m;
    }
    sort(a+1,a+n+1,cmp);
    double sum = 0;
    int pos = 1;
    while(T>0)
    {
        double temp=min(a[pos].m,T);
        T-=temp;
        sum+=temp*a[pos].avg;
        pos++;
    }
    cout << fixed << setprecision(2) << sum;
    return 0;
}

0

hetao3097453 LV 8 @ 2023-5-13 10:46:56

P1028 金币 Lv.1

题目描述

阿里巴巴走进了装满宝藏的藏宝洞。藏宝洞里面有 N 堆金币，第 i 堆金币的总重量和总价值分别是 mi,vi。阿里巴巴有一个承重量为 T 的背包，但并不一定有办法将全部的金币都装进去。他想装走尽可能多价值的金币。所有金币都可以随意分割，分割完的金币重量价值比（也就是单位价格）不变。请问阿里巴巴最多可以拿走多少价值的金币？

思路

对标课上作业。将单位价值求出来，并按照单位价值从大到小排序，加入到背包中。

结构体：

struct Coin
{
    double m,v,avg;
}a[101];

比较：

bool cmp(Coin a,Coin b)
{
    return a.avg > b.avg;
}

参考代码：

#include <iostream>//hetao3097453
#include <algorithm>
#include <iomanip>
using namespace std;
int n;
double t;
struct Coin
{
    double m,v,avg;
}a[101];
bool cmp(Coin a,Coin b)
{
    return a.avg > b.avg;
}
int main()
{
    cin >> n >> t;
    for(int i = 0;i < n;i++)
    {
        cin >> a[i].m >> a[i].v;
        a[i].avg = a[i].v / a[i].m;
    }
    sort(a,a + n,cmp);
    int pos = 0;
    double sum = 0.0;
    while(t > 0)
    {
        double temp = min(a[pos].m,t);
        sum += a[pos].avg * temp;
        t -= temp;
        pos++;
    }
    cout << fixed << setprecision(2) << sum << endl;
    return 0;
}

hetao3097453

2023年5月13日

hetao3097453 @ 2023-5-13 10:59:12

L10-3

-1

hetao13267928 LV 4 @ 2024-4-21 15:30:34

哪错了？

#include<iostream> #include<algorithm> #include <iomanip> using namespace std; int n,t; struct Jinbi { int m,v; double avg; }j[105]; bool cmp(Jinbi x,Jinbi y) { return x.avg>y.avg; } int main() { cin>>n>>t; for(int i=1;i<=n;i++) { cin>>j[i].m>>j[i].v; j[i].avg=j[i].v1.00/j[i].m; } sort(j+n,j+n+1,cmp); int pos=0; double sum=0; while(t>0) { int ans=min(t,j[pos].m); t-=ans; sum+=ansj[pos].avg; pos++; } cout<<fixed<<setprecision(2)<<sum; return 0; }

hetao857718 @ 2024-5-18 17:31:25

m和v用double定义就不会有问题
hetao857718 @ 2024-5-18 17:32:19

那个“.00”加了没用，还报错。
hetao857718 @ 2024-5-18 17:33:59

我为了整你的代码废了半天劲，下次记得加 '''

-1

hetao7177500 LV 3 @ 2024-3-31 16:52:12

作业好吧😄

-2

hetao4582353 LV 8 @ 2023-9-24 12:31:39

这道题和核桃上的一模一样，所以决定不做了，做的意义不大。同意的点个赞。作业就是题解

16 条题解

`金币Lv.1`

Jntm,给个赞吧

P1028 金币 Lv.1

题目描述

思路

参考代码：

金币 Lv.1

信息

状态

开发

支持

16 条题解

金币Lv.1

Jntm,给个赞吧

P1028 金币 Lv.1

题目描述

思路

参考代码：

金币 Lv.1

信息

状态

开发

支持

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`金币Lv.1`