解题时候注意的小问题--------数组开小了（导致超时）

FatMouse' Trade

HDU - 1009

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

下面附上代码：

#include<cstdio>
#include<algorithm>
using namespace std;
struct my
{
    double a,b,c;
}mp[1000];
bool cmp(my x,my y)
{
    return x.c>y.c;
}
int main()
{
    int n,m;
    while(~scanf("%d%d",&m,&n) && (m!=-1 || n!=-1))
    {
        double sum=0;
        for(int i=0;i<n;i++)
        {
            scanf("%lf%lf",&mp[i].a,&mp[i].b);
            mp[i].c=mp[i].a/mp[i].b;
        }
        sort(mp,mp+n,cmp);
        for(int i=0;i<n;i++)
        {
            if(m>mp[i].b)
            {
                sum+=mp[i].a;
                m-=mp[i].b;
            }
            else
            {
                sum+=mp[i].c*m;
                break;
            }
        }
        printf("%.3f\n",sum);
    }
    return 0;
}

//第一次测试  超时 ，，很尴尬，算法是没有问题的，后来才知道是因为数组开小了

//做此记录，数组尽量开大，否则很无奈