Fatmouse ' Trade hdu1009

Title Address: http://acm.hdu.edu.cn/showproblem.php?pid=1009

Fatmouse ' Trade

Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 66570 Accepted Submission (s): 22635

Problem Descriptionfatmouse prepared M pounds of cat food, ready-to-trade with the cats guarding the warehouse containing He favorite food, JavaBean.
The warehouse has N rooms. The i-th contains j[i] pounds of JavaBeans and requires f[i] pounds of cat food. Fatmouse does not has the to trade for all the JavaBeans in the the the same, instead, he may get j[i]* a% of pounds JavaBeans if he Pays f[i]* a% pounds of cat food. Here A is a real number. Now he's assigning this homework to you:tell him the maximum amount of JavaBeans he can obtain. Kankan prepared M Catty of food, prepared with the House long Exchange philosophy treasure ♂ has N rooms. The first room has j[i] ♂ need to f[i] Jin's food. Kankan can not change the entire room ♂, he can use f[i]* a% food for j[i]* a% ♂ now give you a real number M ask you Kankan how much can get ♂

Inputthe input consists of multiple test cases. Each test case is begins with a line containing the non-negative integers M and N. Then N lines follow, each contains, non-negative integers j[i] and f[i] respectively. The last test case was followed by Two-1 ' s. All integers is not greater than 1000.

Outputfor each test case, print in a single line a real number accurate up to 3 decimal places, which is the maximum Amoun T of JavaBeans that Fatmouse can obtain.

Sample INPUT5 37 24 35 220 325 1824 1515 10-1-1

Sample Output13.33331.500

Authorchen, Yue thought: 1 struct structure storage data, 2 by "can use f[i]* a% food for j[i]* a% of ♂", F[i]/j[i] The greater, Kankan from the room for ♂ more,           So use the sort function by F[i]/j[i] from large to small sort, 3 Kankan food is limited, so into the room may change all the ♂, may also press "f[i]* a% food for j[i]* a%" The Law of the ♂ The size of the food and the f[i] of the room should be judged before the Kankan enter the room. The code is as follows:

#include <cstdio>#include<algorithm>using namespacestd;structhuan{Doublej,f; DoubleC;} huan[3005];BOOLCMP (Huan A, Huan B) {returnA.C >B.C;}intMain () {Doublem; intn,t;  while(SCANF ("%lf%d", &m, &n)!=eof && (m!=-1|| n!=-1 ) )    {        intI, j, K; Doublesum=0;  for(i=0; i<n; i++) {scanf ("%LF%LF", &AMP;HUAN[I].J, &huan[i].f); HUAN[I].C= HUAN[I].J/huan[i].f; } sort (Huan, Huan+n, CMP);//sort by F from small to large         for(i=0; i<n; i++)        {            if(M >= huan[i].f)//the whole change.{sum + = HUAN[I].J; M-=huan[i].f; }               Else//The number of rice is not enough to change{sum + = m*huan[i].c; Break; }} printf ("%.3lf\n", sum); }    return 0;}

Fatmouse ' Trade hdu1009