FatMouse' Trade
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 58364 Accepted Submission(s): 19546
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
解体思路:求出交换l率,按交换率从大到小的顺序交换。
代码如下:
#include<stdio.h>
#include<algorithm>
#define INF 0x3f3f3f3f
using namespace std;
struct stu{
double j;
double f;
double v;
};
stu num[1000+10];
int cmp(stu a,stu b){
return a.v>b.v;
}
int main(){
int m,n;
while(scanf("%d%d",&m,&n)){
if(n==-1&&m==-1)break;
for(int i=0;i<n;i++){
scanf("%lf%lf",&num[i].j,&num[i].f);
if(num[i].f==0)num[i].v=INF;
else num[i].v=num[i].j/num[i].f;
}
sort(num,num+n,cmp);
double res,sum;
res=m;sum=0;
for(int i=0;i<n;i++){
if(res>=num[i].f){
res-=num[i].f;
sum+=num[i].j;
}
else if(res<=0)break;
else {
sum+=res*num[i].v;
res=0;
}
}
printf("%.3lf\n",sum);
}
return 0;
}