【学习笔记】〖九度OJ〗题目1433：FatMouse

题目1433：FatMouse

时间限制：1 秒

内存限制：128 兆

特殊判题：否

提交：888

解决：400

题目描述：

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.

输入：

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.

输出：

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.

样例输入：

5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1

样例输出：

13.333
31.500

//简单的贪心策略

#include<iostream>
#include<algorithm>
#include<iomanip>
using namespace std;

class House
{
public:
int weight;
int cost;
double ratio;
void init(int w, int c)
{
weight = w;
cost = c;
ratio = (double)w/(double)c;
}

bool operator < (House h)const
{
return ratio - h.ratio > 0.0001;
}
};

House wh[1001];

int main()
{
freopen("input.in", "r", stdin);
freopen("output.out", "w", stdout);
int m, n, w, c, i;

double res;

while (cin >> m && m!= -1)
{
cin >> n;
for (i=0; i<n; i++)
{
cin >> w >> c;
wh[i].init(w, c);
}

sort(wh, wh+n);

//计算
res = 0;
i=0;
while (m>0 && i<n)
{
if (m>wh[i].cost)
{
res += wh[i].weight;
m -= wh[i].cost;
}
else
{
res += (double)m/(double)wh[i].cost * (double)wh[i].weight;
m = 0;
break;
}
i++;
}
cout << setiosflags(ios::fixed) << setprecision(3);
cout << res << endl;

}

return 0;
}