杭电 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
题目大意:
老鼠有M磅猫食。有N个房间,每个房间前有一只猫,房间里有老鼠最喜欢的食品JavaBean,J[i]。若要引开猫,必须付出相应的猫食F[i]。当然这只老鼠没必要每次都付出所有的F[i]。若它付出F[i]的a%,则得到J[i]的a%。求老鼠能吃到的做多的JavaBean。
1 #include<cstdio> 2 #include<algorithm> 3 using namespace std; 4 struct stu 5 { 6 double a,b; //不用float 7 }a[50000]; 8 bool cmp(stu a,stu b) 9 { 10 return (a.a/a.b)>(b.a/b.b); 11 } 12 int main() 13 { 14 int m,n,i,j; 15 double sum; 16 while(scanf("%d %d",&m,&n)&&(m!=-1)&&(n!=-1)) 17 { 18 sum=0; 19 for(i=0; i < n; i++) 20 { 21 scanf("%lf %lf",&a[i].a,&a[i].b); 22 } 23 sort(a,a+n,cmp); 24 for(i=0;i<n;i++) 25 { 26 if(m > a[i].b) 27 { 28 sum+=a[i].a; 29 m-=a[i].b; 30 } 31 else 32 { 33 sum+=((m/a[i].b)*a[i].a); 34 break; 35 } 36 } 37 printf("%.3lf\n",sum); 38 } 39 }
——将来的你会感谢现在努力的自己。
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