HDOJ1009 贪心水题

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 80550    Accepted Submission(s): 27846

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

 

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买东西都是从来买性价比最高的，这就是人的本性。这题的思路就是这么来的。计算一下性价比，排序后取最高的。

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