FatMouse' Trade

D - FatMouse' Trade

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 1009

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
标准贪心问题，排序相比较就可以了
#include<stdio.h>#include<stdlib.h>#include <algorithm>using namespace std;typedef struct A{    double j,f;    double r;}A;double temp(A a,A b){if(a.r>b.r)return 1;elsereturn 0;}int main(){A  a[1000];A  tmp;  int i,m,b,d,k;double n,sum;while(scanf("%lf%d",&n,&m)!=EOF&&n!=-1&&m!=-1){for(i=0;i<m;i++){        scanf("%lf%lf",&a[i].j,&a[i].f);a[i].r=(double)a[i].j/a[i].f;}sort(a,a+m,temp);  sum=0.0;  for(k=0;k<m&&n>0;k++){if(n>a[k].f){sum+=a[k].j;n-=a[k].f;}else{sum+=a[k].r*n;n=0;}}printf("%.3lf\n",sum);}}