HDU-1009 FatMouse' Trade

链接：http://acm.hdu.edu.cn/showproblem.php?pid=1009

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 31187    Accepted Submission(s): 10074

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 37 24 35 220 325 1824 1515 10-1 -1

 

Sample Output

13.33331.500

 

Author

CHEN, Yue

 

Source

ZJCPC2004

 

Recommend

JGShining

解题报告：贪心思想，属于简单题。7/2=3.5,5/2=2.5,4/3=1.3,然后对结构体排序结果7+5+1*4/3=13.333.

#include<cstdio>#include<algorithm>using namespace std;struct Node{int a,b;double div;}node[1005];int cmp(const Node &x,const Node &y){return x.div>y.div;}int main(){int n,m;double ans;while(scanf("%d%d",&n,&m)&&(n!=-1||m!=-1)){ans=0;for(int i=0;i<m;i++){scanf("%d%d",&node[i].a,&node[i].b );node[i].div=1.0 * node[i].a / node[i].b;//printf("%d %d  %lf \n",node[j].a ,node[j].b , node[j].div );}sort(node,node+m,cmp);//for(int i=0;i<m;i++)//printf("%lf  ", node[i].div );for(int i=0;i<m;i++){if(n>node[i].b){n=n-node[i].b;ans+=node[i].a;}else{//printf("n=%d\n",n);ans+=n*(1.0*node[i].a / node[i].b);break;}}printf("%.3lf\n",ans);}return 0;}