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
思路:按照性价比排好序之后,然后看是否能得到即可
代码:
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
struct node
{
int a,b;
double val;
}p[10005];
bool cmp(node x,node y)
{
return x.val>y.val;
}
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)!=EOF)
{
if(n==-1&&m==-1)
{
break;
}
for(int t=0;t<m;t++)
{
scanf("%d%d",&p[t].a,&p[t].b);
p[t].val=p[t].a*1.0/(p[t].b*1.0);
}
sort(p,p+m,cmp);
double s1=0;
for(int t=0;t<m;t++)
{
if(n>=p[t].b)
{
n-=p[t].b;
s1+=p[t].a;
}
else
{
s1+=(p[t].a*1.0/p[t].b)*n;
break;
}
}
printf("%.3f
",s1);
}
return 0;
}