求解一道ACM的题

[yixiayizi]

题目：
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

我的程序：

#include<iostream>
using namespace std;

typedef struct node
{
	int f,j;
	float info;
}node;

int main()
{
	int m,n,i,j;
	node st[2000],temp;
	float result;
	while(cin>>m>>n&&m!=-1&&n!=-1)
	{
		result=0;
		for(i=0;i<n;i++)
		{
			cin>>st[i].f>>st[i].j;
			if(st[i].j==0)
				st[i].info=1000000000;
			else
				st[i].info=float(st[i].f)/float(st[i].j);
		}
		for(i=0;i<n;i++)
			for(j=0;j<n-1-i;j++)
			{
				if(st[j].info<st[j+1].info)
				{
					temp=st[j];
					st[j]=st[j+1];
					st[j+1]=temp;
				}
			}
		i=0;
		while(i<n)
		{
			if(m>=st[i].j)
			{
				result=result+st[i].f;
				m=m-st[i].j;
				i++;
			}
			else
			{
				result=result+st[i].info*m;
				break;
			}
		}
		printf("%0.3f\n",result);
	}
	return 0;
}

通不过测试，显示是：wrong answer
求助错哪里了，或者哪里考虑不周全了

[yixiayizi]

把float类型都改成了double类型，就都过了。汗死~~~~

st[i].info=1000000000;
这个整数不是最大了吗，你的意思是换成长整型吗？

[zhangpingfly]

这个用的就是贪心~~~
下面是我在杭电ACM上的answer

#include <iostream>
#include <stdio.h>
#include <string>
using namespace std;
int main()
{
    int m,n,i,j,num;
    int residual,a[2002];
    double max,sum,b[1000];
    while (cin>>m>>n)
    {
        sum=0;residual=m;num=n;
        if(m==-1&&n==-1)exit(0);
        for (i=0;i<n*2;i++)
            scanf("%d",&a[i]);
        for(i=0,j=0;i<n;i++,j=j+2)
            b[i]=(double)a[j]/a[j+1];
        while (residual>0&&num>0)
        {
            max=b[0];j=0;
            for (i=1;i<n;i++)
            {
                if(b[i]>max)
                {
                    max=b[i];
                    j=i;
                }
            }
            if(residual>=a[j*2+1])
            {
                residual-=a[j*2+1];
                sum+=a[j*2];
                b[j]=0;
            }
            else
            {
                sum=sum+(double)residual*a[j*2]/a[j*2+1];
                residual=0;
            }
            num--;
        }
        printf("%.3f\n",sum);
    }
    return 0;
}

[visame]

经典的贪心算法。

#include <iostream>
using namespace std;

void Merge(double *Cat,double *Mouse,int start,int middle,int end)
{
	double *TempCat=new double [end-start+1];
	double *TempMouse=new double [end-start+1];
	int i=start,j=middle+1,marker=0;
	while (i<=middle&&j<=end)
	{
		while (Cat[i]/Mouse[i]<=Cat[j]/Mouse[j]&&i<=middle&&j<=end)
		{
			TempCat[marker]=Cat[i];
			TempMouse[marker]=Mouse[i];
			++marker;
			++i;
		}
		while (Cat[j]/Mouse[j]<=Cat[i]/Mouse[i]&&i<=middle&&j<=end)
		{
			TempCat[marker]=Cat[j];
			TempMouse[marker]=Mouse[j];
			++marker;
			++j;
		}
	}
	while (i<=middle)
	{
		TempCat[marker]=Cat[i];
		TempMouse[marker]=Mouse[i];
		++marker;
		++i;
	}
	while (j<=end)
	{
		TempCat[marker]=Cat[j];
		TempMouse[marker]=Mouse[j];
		++marker;
		++j;
	}
	for (int m=start;m<=end;++m)
	{
		Cat[m]=TempCat[m-start];
		Mouse[m]=TempMouse[m-start];
	}
	delete []TempCat;
	delete []TempMouse;
}
void MergeSort(double *Cat,double *Mouse,int start,int end)
{
	if (start<end)
	{
		int middle=(start+end)/2;
		MergeSort(Cat,Mouse,start,middle);
		MergeSort(Cat,Mouse,middle+1,end);
		Merge(Cat,Mouse,start,middle,end);
	}
}

int main()
{
	int N;
	double M;
	while(scanf("%lf%d",&M,&N)!=EOF&&N!=-1)
	{
		double *Cat=new double [N];
		double *Mouse=new double [N];
		for (int i=0;i<=N-1;++i)
		{
			scanf("%lf%lf",&Mouse[i],&Cat[i]);
		}
		MergeSort(Cat,Mouse,0,N-1);
		int j=0;
		double Result=0;
		for (j=0;j<=N-1;++j)
		{
			if (M-Cat[j]>=0)
			{
				M=M-Cat[j];
				Result+=Mouse[j];
			}
			else
			{
				Result+=M*Mouse[j]/Cat[j];
				printf("%.3f\n",Result);
				break;
			}
		}
	}
}

[visame]

看了一遍，思路很清晰。比我的还好，不过我主要是练习一下合并排序。
应该是浙江的。
可能出错的两个地方：

            if(st[i].j==0)
                st[i].info=1000000000;//此数太小，换成更大的，测试数据一般超级反常
            else
                st[i].info=float(st[i].f)/float(st[i].j);//这里把float换成double提高精度

Try again!Waiting for your good news!

[visame]

又想了一会儿，肯定是

存在j=0的数据，而你又图简单把st[i].info=1000000000;造成结果偏大。
比较的时候就要多写两句而已。

result=result+st[i].info*m;//here，最后应该不增加，但你的增加了
break;

[zac372424332]

我也弄了好久没弄出来，后来把float改成double就行了，谢谢楼主。