FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 66662    Accepted Submission(s): 22664

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

 

Sample Output

13.33331.500

 

Author

CHEN, Yue

 

Source

ZJCPC2004

 

Recommend

JGShining

 

题意就是说有n种换取方法一共有m的粮食最多可以换多少

简单的贪心按照单位粮食换取的多少排序然后由大到小取就好了

1234567891011121314151617181920212223242526272829303132333435363738

#include<cstdio>#include<algorithm>using namespace std;struct node{double j;double f;double cha;}a[1011];bool cmp(node a,node b){return a.cha>b.cha;}int m;int n;int main(){while(scanf("%d %d",&m,&n)){if(m==-1&&n==-1)break;for(int i=0;i<n;i++){scanf("%lf %lf",&a[i].j,&a[i].f);a[i].cha=a[i].j/a[i].f;//算出单价}sort(a,a+n,cmp);//排序double sum=0;int i=0;for(int i=0;i<n;i++){if(m>a[i].f){//判断剩余的钱是否可以把这个仓库的全部换走sum+=a[i].j;m-=a[i].f;}else{//不能的话就换取可以换到的sum+=a[i].cha*m;break;}}printf("%.3lf\n",sum);}return 0;}