A - A

Time limit : 1 s Memory limit : 32 mb
Submitted : 6 Accepted : 2
64bit Integer Format : %lld

 

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

struct info{
    double val;
    int pos;
};
int amount(info a[],int n)
{
    int i,j;
    for(i=0;i<n;i++)
    {
        for(j=i+1;j<n;j++)
        {
           if(a[j].val>a[i].val)
           {
               double t;
               t = a[i].val;
               a[i].val=a[j].val;
               a[j].val=t;
               int tt;
               tt=a[i].pos;
               a[i].pos=a[j].pos;
               a[j].pos=tt;
           }
        }
    }
    return 0;
}
int main()
{

    int M,N;
    int i;
    while(1)
    {
        scanf("%d %d",&M,&N);
        if(M==-1&&N==-1)
            break;
        info price[N]={0};
        int F[N]={0},J[N]={0};
        for(i=0;i<N;i++)
        {
            scanf("%d %d",&J[i],&F[i]);
            price[i].val=(double)J[i]/F[i];
            price[i].pos=i;
        }
        amount(price,N);
        double output=0;
        for(i=0;i<N;i++)
        {
            if(M>=F[price[i].pos])
            {
                output+=price[i].val*F[price[i].pos];
                M-=F[price[i].pos];
            }
            else if(M>0&&M<F[price[i].pos])
            {
                output+=price[i].val*M;
                M=0;
                break;
            }
        }
        printf("%.3lf\n",output);
    }
    return 0;
}

然而在oj里面。。。Runtime Error

没办法啊，只好看看哪里可以改进

2.sort

#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;
}

key3：without sort:

#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;
}