HDU 1009：FatMouse' Trade（简单贪心）

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 41982    Accepted Submission(s): 13962

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

  

题意就是老鼠用猫粮换鼠粮。。。（Orz）。。求它最多能换多少。。一道贪心水题。。

#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<iostream>#include<vector>#include<queue>#include<cmath>#define f1(i, n) for(int i=0; i<n; i++)#define f2(i, n) for(int i=1; i<=n; i++)using namespace std;const int M = 1005;int n, m;double ans;double t;struct node{    double J;    double F;    double c;}Q[M];int cmp (node a, node b){    return a.c > b.c;}int main(){    while(scanf("%d%d", &n, &m)!=EOF)    {        ans = 0;        t = (double) n;        memset(Q, 0, sizeof(Q));        if(n==-1 && m==-1)            break;        f1(i, m)           {               scanf("%lf%lf", &Q[i].J, &Q[i].F);               Q[i].c = Q[i].J / Q[i].F;           }        sort(Q, Q+m, cmp);        f1(i, m)        {            if( Q[i].F<=t )                {                    ans += Q[i].J;                     t -= Q[i].F;                }            else            {                ans += t * Q[i].c;                break;            }        }        printf("%.3lf\n", ans);    }    return 0;}