HDU 1009 FatMouse' Trade(贪心)
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
1 #include<stdio.h> 2 #include<algorithm> 3 using namespace std; 4 int m,n; 5 struct x{ 6 int j,c; 7 double q; 8 }a[1010]; 9 bool cmp(x a,x b){//贪心,按照最少的猫粮换最多的老鼠吃的排序 10 return a.q>b.q; 11 } 12 13 int main(){ 14 while(scanf("%d%d",&m,&n)!=EOF){ 15 if(m==-1&&n==-1){ 16 break; 17 } 18 for(int i=1;i<=n;i++){ 19 scanf("%d%d",&a[i].j,&a[i].c); 20 a[i].q=a[i].j/(1.0*a[i].c);//求出每组的转化率 21 } //乘以0.1,转化为浮点型 22 sort(a+1,a+1+n,cmp); 23 double sum=0; 24 for(int i=1;i<=n;i++){ 25 if(m>=a[i].c){//如果老鼠手里有的猫粮比要求的多,把所有的老鼠吃的都换来 26 sum+=a[i].j;//加上换来的鼠粮 27 m-=a[i].c;//手里剩下的猫粮 28 } 29 else{ 30 sum+=a[i].q*m;//不足,就按比例换 31 m-=m; 32 } 33 if(m==0){ 34 break; 35 } 36 } 37 printf("%.3f\n",sum); 38 } 39 }
(寒いですね)
posted on 2018-12-30 16:13 甜甜圈不懂巧克力的苦 阅读(126) 评论(0) 编辑 收藏 举报