Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 43156 Accepted Submission(s): 14417
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
Recommend
JGShining
贪心.
贪心策略是根据价值比进行降序排序 :J[i]/F[i];
代码如下:
1 #include <cstdio>
2 #include <cstring>
3 #include <iostream>
4 #include <algorithm>
5 using namespace std;
6 #define MAX 1001
7 struct node
8 {
9 int f;
10 int j;
11 double p;
12 };
13 struct node x[MAX];
14 int n,m;
15 bool cmp(struct node x,struct node y)
16 {
17 if(x.p==y.p) return x.f<y.f;
18 return x.p>y.p;
19 }
20 void out()
21 {
22 for(int i=0;i<n;i++)
23 cout<<x[i].p<<" ";
24 cout<<endl;
25 }
26 void init()
27 {
28 memset(x,0,sizeof(x));
29 }
30 void read()
31 {
32 int i;
33 for(i=0;i<n;i++)
34 {
35 scanf("%d %d",&x[i].j,&x[i].f);
36 x[i].p=(double)(x[i].j*1.0/(x[i].f*1.0));
37 }
38 sort(x,x+n,cmp);
39 }
40 void cal()
41 {
42 int i;
43 double res=m;
44 double ans=0;
45 for(i=0;i<n;i++)
46 {
47 if(res>=x[i].f)
48 {
49 res-=x[i].f;
50 ans+=x[i].j;
51 }
52 else
53 {
54 ans=ans+res*x[i].p;
55 res-=res;
56 }
57 }
58 printf("%.3f\n",ans);
59 }
60 void solve()
61 {
62 init();
63 read();
64 cal();
65 }
66
67 int main()
68 {
69 while(scanf("%d %d",&m,&n)!=EOF)
70 {
71 if(n==-1&&m==-1) break;
72 solve();
73 }
74 return 0;
75 }