hdu 1009 FatMouse' Trade
1 #include<stdio.h> 2 #include<stdlib.h> 3 #include<algorithm> 4 using namespace std; 5 const int MAXN = 1010; 6 struct node 7 { 8 double j,f; 9 double r; 10 }a[MAXN]; 11 bool cmp(node a,node b) 12 { 13 return a.r > b.r; 14 } 15 int main() 16 { 17 int N; 18 double M; 19 double ans; 20 while(scanf("%lf%d",&M,&N)) 21 { 22 if(M==-1&&N==-1) break; 23 for(int i=0;i<N;i++) 24 { 25 scanf("%lf%lf",&a[i].j,&a[i].f); 26 a[i].r=(double)a[i].j/a[i].f; 27 } 28 sort(a,a+N,cmp); 29 ans=0; 30 for(int i=0;i<N;i++) 31 { 32 if(M>=a[i].f) 33 { 34 ans+=a[i].j; 35 M-=a[i].f; 36 } 37 else 38 { 39 ans+=(a[i].j/a[i].f)*M; 40 break; 41 } 42 } 43 printf("%.3lf\n",ans); 44 } 45 return 0; 46 }
FatMouse' TradeTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 80808 Accepted Submission(s): 27951 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. 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
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作者:红雨
出处:https://www.cnblogs.com/52why
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