SDJZU_新生_贪心_FatMouse' Trade

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昨天就是练练手，培训今天才开始

SDJZU_新生_贪心

27:25:42

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Current Time: 2015-01-24 18:55:42Contest Type: PrivateStart Time: 2015-01-23 15:30:00Contest Status: RunningEnd Time: 2015-01-24 21:30:00Manager: ACboy

nickname改成自己真实姓名

C - FatMouse' Trade

Crawling in process...Crawling failedTime Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

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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

#include<stdio.h>#include<algorithm>using namespace std;struct A{    int j,f;    double z;}a[1007];int cmp(A a,A b){    return a.z>b.z;}int main(){    int M,N;    while(scanf("%d%d",&M,&N)!=EOF)    {        if(M==-1&&N==-1)        {            break;        }        A a[1007];        int i;        for(i=0;i<N;i++)        {            scanf("%d%d",&a[i].j,&a[i].f);            a[i].z=((double)a[i].j/a[i].f);        }        sort(a,a+N,cmp);        double b=0;        for(i=0;M>0&&i<N;i++)        {            if(M>=a[i].f)            {                M-=a[i].f;                b+=a[i].j;            }            else            {                b+=(((double)M/a[i].f)*a[i].j);                M=0;            }        }        printf("%.3f\n",b);    }    return 0;}

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