杭电1009-FatMouse' Trade（贪心）

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 54937    Accepted Submission(s): 18422

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

 

这是个简单贪心，先计算出那种交易最值得，就先进行那个，直到自己的资金为0

AC代码：

#include<cstdio>#include<algorithm>using namespace std;struct node {int v,w;double q;}s[10000];bool cmp(node x,node y){return x.q>y.q;}int main(){int i,n,m;double a[10000];while(scanf("%d%d",&m,&n),m!=-1||n!=-1){for(i=0;i<n;i++){scanf("%d%d",&s[i].v,&s[i].w);s[i].q=s[i].v*1.0/s[i].w;}sort(s,s+n,cmp);double sum=0;for(i=0;i<n;i++){if(s[i].w<=m){sum+=s[i].v;m-=s[i].w;}else{sum+=m*s[i].q;break;}}printf("%.3lf\n",sum);}return 0; }