FatMouse' Trade
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 33038 Accepted Submission(s): 10693
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
Author
CHEN, Yue
Source
ZJCPC2004
Recommend
JGShining
1 #include<cstdio> 2 #include<cmath> 3 #include<iostream> 4 #include<algorithm> 5 #include<cstring> 6 using namespace std; 7 8 struct node 9 { 10 int ja,va; 11 double sor; 12 }edge[10001]; 13 14 bool cmp(node a,node b) 15 { 16 return a.sor<b.sor; 17 } 18 19 int main() 20 { 21 int m,n; 22 while(scanf("%d%d",&m,&n)!=EOF) 23 { 24 if(m==-1&&n==-1)break; 25 int i; 26 for(i=0;i<n;i++) 27 { 28 scanf("%d%d",&edge[i].ja,&edge[i].va); 29 edge[i].sor=edge[i].va*1.0/edge[i].ja; 30 } 31 32 sort(edge,edge+n,cmp); 33 34 double sum=0; 35 for(i=0;i<n;i++) 36 { 37 if(m>edge[i].va) 38 { 39 sum+=edge[i].ja; 40 m-=edge[i].va; 41 } 42 else 43 { 44 sum+=m*1.0/edge[i].va*edge[i].ja; 45 break; 46 } 47 } 48 printf("%0.3lf ",sum); 49 } 50 return 0; 51 }