FatMouse' Trade

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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 37 24 35 220 325 1824 1515 10-1 -1

Sample Output

 13.33331.500

这个问题和背包问题几乎一样，只是有点细微差别。

要注意的是，有可能某个房间里有JavaBean但是需要的猫粮是零，这种情况要考虑。

代码：

#include<stdio.h>main(){int i,j;int a[1100],b[1100],tempx;double sum;double c[1100],temp;int m,n;while(scanf("%d%d",&m,&n)&&(m!=-1&&n!=-1)){sum=0;for(i=0;i<n;i++){scanf("%d%d",&a[i],&b[i]);}for(i=0;i<n;i++){if(b[i]==0){sum=sum+a[i];for(j=i;j<n-1;j++){a[j]=a[j+1];b[j]=b[j+1];}n--;i--;}}for(i=0;i<n;i++){c[i]=1.0*a[i]/b[i];}for(i=0;i<n;i++){for(j=0;j<n-1;j++){if(c[j]<c[j+1]){temp=c[j];c[j]=c[j+1];c[j+1]=temp;tempx=a[j];a[j]=a[j+1];a[j+1]=tempx;tempx=b[j];b[j]=b[j+1];b[j+1]=tempx;}}}for(i=0;i<n;i++){if(m>b[i]){sum=sum+a[i];m=m-b[i];}else{sum=sum+m*c[i];                break;}} printf("%.3lf\n",sum);}}

还有同学的一个：

#include<stdio.h>#include<algorithm>using namespace std;struct node{double a;double b;double c;}s[1000];bool cmp(node a,node b){return a.c>b.c;}int main(){    double p,q;   int i,m,n;  while(scanf("%d%d",&m,&n),(m!=-1)&&(n!=-1)){ p=0; q=0; for(i=0;i<n;i++){scanf("%lf%lf",&s[i].a,&s[i].b);s[i].c=s[i].a/s[i].b;}sort(s,s+n,cmp);i=0;while(p<m&&i<n){p+=s[i].b;q+=s[i].a;i++;}printf("%.3f\n",q-(p-m)*(s[i-1].a/s[i-1].b));}}

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