FatMouse’ Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20968    Accepted Submission(s): 6501

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 3 7 2 4 3 5 2 20 3 25 18 24 15 15 10 -1 -1  

Sample Output 13.333 31.500  

Author CHEN, Yue  

Source
ZJCPC2004  

Recommend JGShining     很简单的贪心题， 先按照价值/代价的比值来排序，肯定是先买比值大的。 代码如下：

#include<stdio.h>
#include<stdlib.h>

const int MAXN = 1010;
struct node
{
    double j,f;
    double r;
}a[MAXN];
int cmp(const void *a,const void *b)//从大到小排序 
{
    struct node *c=(node *)a;
    struct node *d=(node *)b;
    if(c->r > d->r) return -1;
    else return 1;
}    
int main()
{
    int N;
    double M;
    double ans;
    while(scanf("%lf%d",&M,&N))
    {
        if(M==-1&&N==-1) break;
        for(int i=0;i<N;i++)
        {
           scanf("%lf%lf",&a[i].j,&a[i].f);
           a[i].r=(double)a[i].j/a[i].f;
        }    
        qsort(a,N,sizeof(a[0]),cmp);
        ans=0;
        for(int i=0;i<N;i++)
        {
            if(M>=a[i].f)
            {
                ans+=a[i].j;
                M-=a[i].f;
            }    
            else 
            {
                ans+=(a[i].j/a[i].f)*M;
                break;
            }    
        }   
        printf("%.3lf\n",ans); 
    }    
    return 0;
}    

  sort排序版本：

#include<stdio.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
const int MAXN = 1010;
struct node
{
    double j,f;
    double r;
}a[MAXN];
/*
int cmp(const void *a,const void *b)//从大到小排序 
{
    struct node *c=(node *)a;
    struct node *d=(node *)b;
    if(c->r > d->r) return -1;
    else return 1;
}    */
bool cmp(node a,node b)
{
    return a.r  >  b.r;
}    
int main()
{
    int N;
    double M;
    double ans;
    while(scanf("%lf%d",&M,&N))
    {
        if(M==-1&&N==-1) break;
        for(int i=0;i<N;i++)
        {
           scanf("%lf%lf",&a[i].j,&a[i].f);
           a[i].r=(double)a[i].j/a[i].f;
        }    
        //qsort(a,N,sizeof(a[0]),cmp);
        sort(a,a+N,cmp);
        ans=0;
        for(int i=0;i<N;i++)
        {
            if(M>=a[i].f)
            {
                ans+=a[i].j;
                M-=a[i].f;
            }    
            else 
            {
                ans+=(a[i].j/a[i].f)*M;
                break;
            }    
        }   
        printf("%.3lf\n",ans); 
    }    
    return 0;
}