FatMouse' Trade

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 50 Accepted Submission(s) : 19

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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

Sample Output

13.333
31.500

Author

CHEN, Yue

Source

ZJCPC2004

Statistic | Submit | Back

大意：老鼠有m磅猫粮，在一个地方有n个房间，每个这里有猫在看守，第i个房间里有j[i]镑豆子要求f[i]磅猫粮房间里的豆子便可全部带走，但老鼠不一定要全部带走，如果老鼠给了猫f[i]*a%磅的猫粮，则他可获得j[i]*a%磅的豆子，求老鼠可获得的最大数量的豆子是多少？

思路：问题相当于买东西，如何花最少的钱来买最近可能多的东西。所以那件东西花的钱（猫粮）最少就先买哪一个，也就是说花同样的钱（猫粮），哪个买的多就先买哪个。

即先买j[i]/f[i]最大的那个，若不猫要求的猫粮，则就买m/f[i]*j[i]的豆子。可先将j[i]/f[i]排序，再从最多的往下买

代码：

#include <iostream>
#include<iomanip>
#define MAX 9999999
using namespace std;
double j[1000],f[1000],p[1000];
void order(double p[],double j[],double f[],int n)
{int i,s,k;for(i=n; i>0; i--){k=0;for(s=0; s<i; s++){if(p[k]>=p[s]){k=s;}}double t;t=p[k];p[k]=p[i-1];p[i-1]=t;t=j[k];j[k]=j[i-1];j[i-1]=t;t=f[k];f[k]=f[i-1];f[i-1]=t;}
}
int main()
{int n,i;
double s,m;while(cin>>m>>n&&(n!=-1&&m!=-1)){s=0;for(i=0; i<n; i++){cin>>j[i]>>f[i];if(f[i]!=0)//分母等于零的情况处理p[i]=j[i]/f[i];elsep[i]=MAX;//max假定无穷}order(p,j,f,n);//排序for(i=0; i<n; i++){if(m>=f[i])//足够付得起{s=s+j[i];m=m-f[i];}else if(m>=0&&m<f[i])付不起时{s=s+m*p[i];m=0;}}cout<<setiosflags(ios::fixed)<<setprecision(3);cout<<s<<endl;}return 0;
}

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HDU 1009FatMouse' Trade