题目来源：hdu 1009 FatMouse’ Trade
FatMouse’ Trade Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 54581 Accepted Submission(s): 18299
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 题目大意： 输入m,n，m可以看做总容量，n组数据，每组数据输入J、F；J可看做该物品的总价值，F可看做该物品所占空间为F。所求的是m空间所能装的最大价值是多少，保留三位小数。 题目分析： 此题意在考查贪心思想，想要获得最大价值，我们就要首先取单位价值最大的，将其取完或者取满空间，若空间较大，则再取剩余的单位价值最大的物品。由此可知，此题需要对物品的单位价值进行排序，即(J/F)值，然后使用贪心即可。 AC代码：
;struct node{//用结构体存放该物品的价值、体积、单位价值 int j,f;double v;};node edge[1010];int cmp(node a,node b) //对物品的单位价值从大到小进行排序 {return a.v>b.v;}int main(){int m,n;while(scanf(“%d%d”,&m,&n)!=EOF&&m!=-1&&n!=-1){for(int i=0;i<n;i++)//输入每件物品的价值及体积，并计算单位体积 {scanf(“%d%d”,&edge[i].j,&edge[i].f);edge[i].v=(double)edge[i].j/edge[i].f;}sort(edge,edge+n,cmp); (int i=0;i<n;i++){if(m>=edge[i].f) //若空间较大，则取完此物品，，剩余空间为m减去该物品的体积 {m-=edge[i].f;ans+=edge[i].j; //ans计算当前所取的最大价值 }else{ans+=edge[i].v*m; //空间不足以装完此物品时，就装满剩余空间 break;//此时所装价值为单位价值与所装空间之积，已无法继续装，故跳出 }}printf(“%.3lf\n”,ans);}return 0;}
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