hdoj 1009 FatMouse' Trade解题报告

ACM

发布日期: 2016-02-11

文章字数: 647

阅读时长: 3 分

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FatMouse’ Trade
**Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 60482 Accepted Submission(s): 20335

**

Problem Description

FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean.
The warehouse has N rooms. The i-th room contains J[i] pounds of JavaBeans and requires F[i] pounds of cat food. FatMouse does not have to trade for all the JavaBeans in the room, instead, he may get J[i]* a% pounds of JavaBeans if he pays F[i]* a% pounds of cat food. Here a is a real number. Now he is assigning this homework to you: tell him the maximum amount of JavaBeans he can obtain.

Input

The input consists of multiple test cases. Each test case begins with a line containing two non-negative integers M and N. Then N lines follow, each contains two non-negative integers J[i] and F[i] respectively. The last test case is followed by two -1's. All integers are not greater than 1000.

Output

For each test case, print in a single line a real number accurate up to 3 decimal places, which is the maximum amount of JavaBeans that FatMouse can obtain.

Sample Input

5 3 7 2 4 3 5 2 20 3 25 18 24 15 15 10 -1 -1

Sample Output

13.333 31.500

我的天，一道很简单的贪心题，然而一开始就TE，我还以为循环遍历太多，需要剪枝之类的优化，后来发现了两个地方的错误。。。好丢人Orz，，，

while (scanf("%d %d",&m,&n)&&m!=-1)

一开始是这里，忘记判断m=-1时结束输入，这应该是造成TE的原因；

第二个地方，是关于比例j[i]/f[i]的处理，f[i]可能为0，将其mods设为double类型最大值即可。

scanf("%lf %lf",&a[i].a,&a[i].b);
if(a[i].b==0)
{
    a[i].mods=DBL_MAX;
}
else
{
    a[i].mods=a[i].a/a[i].b;
}

嗯，就这么多，下面是完整代码：

#include<iostream>
#include<stdio.h
#include<math.h>
#include<algorithm
#define DBL_MAX 1.7976931348623158e+308
using namespace std;
struct food
{
    double a,b,mods;
};
food a[1001];
bool compare(food a,food b)
{
    return a.mods<b.mods;
}
int main()
{
    int m,n;
    double ans;
    while (scanf("%d %d",&m,&n)&&m!=-1)
    {
        for (int i = 0; i < n; i += 1)
        {
                scanf("%lf %lf",&a[i].a,&a[i].b);
                if(a[i].b==0)
                {
                    a[i].mods=DBL_MAX;
                }
                else
                {
                    a[i].mods=a[i].a/a[i].b;
                }
        }
        sort(a,a+n,compare);
        ans=0.0;
        while (n)
        {
            n--;
            if (m>=a[n].b)
            {
                m-=a[n].b;
                ans+=a[n].a;
            }
            else
            {
                ans+=m*a[n].mods;
                m=0;
            }
            if(a[n].b!=0&&m==0)break;
        }
        printf("%.3lf\n",ans);
    }
    return 0;
}

细节啊细节，c++中对数的处理和java中还是有些区别的，需要谨慎。

crazyX

http://blog.jingwei.site/2016/02/11/hdoj1009/