HDU1009_FatMouse' Trade【贪心】【水题】
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FatMouse' Trade
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 44470 Accepted Submission(s): 14872
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
题目大意:有N个房间,每个房间存有FatMouse喜欢吃的食物,但是每个房间
的食物都需要用相应的猫粮去换。FatMouse 有M磅的猫粮,为它最多能换到多
少的食物。
思路:贪心方法。用结构体存每间房间的食物量和所需猫粮量。按食物的单价(
即食物/猫粮的大小)进行排列,每次选单价最小的购买,知道M磅猫粮用完
#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;struct warehouse{ double j; double f;}a[1100];bool cmp(warehouse a,warehouse b){ return a.f/a.g < b.f/b.g;}int main(){ int N; double M; while(~scanf("%lf%d",&M,&N)&& (M!=-1||N!=-1)) { memset(a,0,sizeof(a)); for(int i = 0; i < N; i++) { scanf("%lf%lf",&a[i].j,&a[i].f); } sort(a,a+N,cmp); double sum = 0; for(int i = 0; i < N; i++) { if(M <= 0.000001) break; if(M >= a[i].f) { sum += a[i].j; M -= a[i].f; } else { sum += M*a[i].j/a[i].f; M = 0; } } printf("%.3lf\n",sum); } return 0;}
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