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): 30917 Accepted Submission(s): 9975
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 <stdio.h>#include <algorithm>using namespace std;struct Node{ double j,f,p;} node[10000];int cmp(Node x,Node y){ return x.p>y.p;}int main(){ int m,n; while(~scanf("%d%d",&n,&m) && (m!=-1 || n!=-1)) { double sum = 0,max = 0; int i,j; for(i = 0; i<m; i++) { scanf("%lf%lf",&node[i].j,&node[i].f); node[i].p = node[i].j/node[i].f; } sort(node,node+m,cmp); for(i = 0; i<m; i++) { if(n>node[i].f) { sum+=node[i].j; n-=node[i].f; } else { sum+=node[i].p*n; break; } } printf("%.3lf\n",sum); } return 0;}
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