hdu 1009 FatMouse' Trade(贪心)

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 45474    Accepted Submission(s): 15250

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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AC CODE:

#include<iostream>#include<cstdio>#include<algorithm>#include<cstdlib>using namespace std;struct M{    int a, b;    double k;} s[1100];int cmp ( struct M a, struct M b ){    return a.k < b.k;}int main(){    int n, m;    while ( cin >> m >> n )    {        if ( n == -1 && m == -1 )            break;        else        {            for ( int i = 0; i < n; i++ )            {                cin >> s[i].a >> s[i].b;                s[i].k = s[i].a * 1.0 / s[i].b;            }            sort ( s, s + n, cmp );            double max = 0;            for ( int i = n - 1; i >= 0; i-- )            {                if ( m >= s[i].b )                {                    max += s[i].a;                    m -= s[i].b;                }                else if ( m < s[i].b && m > 0 )                {                    max += s[i].a * 1.0 * m / s[i].b;                    m = 0;                }                else if ( m == 0 )                    break;            }            printf ( "%.3f\n", max );        }    }    return 0;}