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
13.333
31.500
简单贪心问题。先利用结构体对J和F进行排序。然后依次累加即可。
#include
#include
#include
using namespace std;
struct Room
{
int j, f;
};
Room room[1005];
bool cmp(const Room a, const Room b)
{
return ((double)a.j / a.f > (double)b.j / b.f);
}
int main()
{
int m, n;
while (cin >> m >> n && m!=-1 && n!=-1)
{
int i;
for (i = 0; i > room[i].j >> room[i].f;
}
sort(room , room + n, cmp);
int jsum=0, fsum = 0;
for (i = 0; i = m) { break; }
}
cout m)
{
fsum -= room[i].f;
jsum -= room[i].j;
double javasum;
javasum = jsum+(double)room[i].j *(double(m - fsum) / room[i].f);
cout
