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CDZSC_2015寒假新人（1）——基础 C
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 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1
Sample Output
13.333
31.500
#include <iostream>#include <cstdio>#include <algorithm>using namespace std;struct laoshu{ int j; int f; double s;};bool cmp(laoshu stud1,laoshu stud2){ return stud1.s>stud2.s;}int main(){#ifdef CDZSC_OFFLINE freopen("in.txt","r",stdin);#endif int i,m,n; double sum; laoshu stud[1010]; while(scanf("%d%d",&m,&n)&&m!=-1&&n!=-1) { sum=0; for(i=0; i<n; i++) { scanf("%d%d",&stud[i].j,&stud[i].f); stud[i].s=1.0*stud[i].j/stud[i].f; } sort(stud,stud+n,cmp); for(i=0; i<n; i++) { if(m>=stud[i].f) { sum+=stud[i].j; m-=stud[i].f; } else { sum+=stud[i].s*m; break; } } printf("%.3lf\n",sum); } return 0;}
CDZSC_2015寒假新人（1）——基础 C
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