Hdu - 1009 - FatMouse' Trade
上题目
FatMouse' Trade
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 33316 Accepted Submission(s): 10813
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.
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
题意就是有m单位猫食,n堆的豆,每堆豆有一个余量和要这一堆豆的代价,每一堆豆可以只要一部分,那么付出的代价就按这一部分占这一堆豆的余量的百分比算。问用这m单位猫食可以最多得到多少豆。
贪心的基础题,白书上也有类似的题目。解法就是得出每一堆豆的单价(余量/代价),先取完单价最高的豆,然后再取第二的···直到猫食用完。
上代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <stdio.h> 2 #include <string.h> 3 #include <algorithm> 4 #define MAX 1000+10 5 using namespace std; 6 7 typedef struct 8 { 9 double J,F; 10 double S; 11 }price; 12 13 bool cmp(price x,price y) {return x.S>y.S;} 14 15 price p[MAX]; 16 17 int main() 18 { 19 int i,m,n; 20 double sum; 21 //freopen("data.txt","r",stdin); 22 while(scanf("%d %d",&m,&n)!=EOF) 23 { 24 if(m==-1 && n==-1) break; 25 memset(p,0,sizeof(p)); 26 for(i=0;i<n;i++) 27 { 28 scanf("%lf %lf",&p[i].J,&p[i].F); 29 p[i].S=p[i].J/p[i].F; 30 } 31 sort(p,p+n,cmp); 32 sum=0; 33 for(i=0;i<n;i++) 34 { 35 if(m>=p[i].F) {m-=p[i].F;sum+=p[i].J;} 36 else {sum+=p[i].J*m/p[i].F;break;} 37 } 38 printf("%.3lf\n",sum); 39 } 40 return 0; 41 }