【HDU 1009】FatMouse' Trade
2016-02-05 16:08
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题
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
题意:给你m克猫粮,n个房间分别可以用猫粮以J[i]/F[i]的比例换取至多J[i]克咖啡豆,问最多能换取多少咖啡豆
分析:先排下序再贪心
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
题意:给你m克猫粮,n个房间分别可以用猫粮以J[i]/F[i]的比例换取至多J[i]克咖啡豆,问最多能换取多少咖啡豆
分析:先排下序再贪心
#include <stdio.h> #include <algorithm> using namespace std; int n,u; double m,ans; struct room { double j,f,p; } r[1005]; int cmp(room a,room b) { return a.p>b.p; } int main() { while(scanf("%lf%d",&m,&n)&&n!=-1&&m!=-1) { for(int i=0;i<n;i++) scanf("%lf%lf",&r[i].j,&r[i].f), r[i].p=r[i].j/r[i].f; sort(r,r+n,cmp); u=ans=0; while(m&&u<n) { if(r[u].f>m) ans+=m*r[u].p,m=0; //剩下的猫粮不够把这个房间的咖啡豆换过来,那就能换多少换多少 else ans+=r[u].j,m-=r[u].f; //够的话,这间房子全部咖啡豆换过来 u++; } // while(m>=r[u].f&&u<n) //换种写法 // { // ans+=r[u].j; // m-=r[u].f; // u++; // } // if(u!=n) ans+=m*r[u].p; printf("%.3lf\n",ans); } return 0; }
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