FatMouse' Trade
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FatMouse' Trade
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 37292 Accepted Submission(s): 12311
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.
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
include<cstdio>#include<algorithm>#include<iostream>using namespace std;struct info{ int j; int f; double ave;}unit[10005];int cmd(const info &x,const info &y){ //printf("fhdshfjks"); return x.ave<y.ave;}int main(){ int i,j,m,n; double sum; //info unit[10005]; while(scanf("%d%d",&m,&n)!=EOF) { if((n==-1)&&(m==-1)) break; for(i=0;i<n;i++) { scanf("%d%d",&unit[i].j,&unit[i].f); unit[i].ave=unit[i].f/(unit[i].j*1.0); } sort(unit,unit+n,cmd); //for(i=0;i<n;i++) //printf("%f\n",unit[i].ave); sum=0; //printf("%d",m); for(i=0;i<n&&m!=0;i++) { if(m>=unit[i].f) { sum=sum+unit[i].j; //printf("hdjks%f\n",sum); m=m-unit[i].f; } else { sum=sum+(m/(unit[i].f*1.0))*unit[i].j; //printf("aaa%f\n",sum); m=0; break; } } printf("%.3f\n",sum); } return 0;}
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