FatMouse' Trade(sourtqsourt)

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FatMouse' Trade

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 15 Accepted Submission(s) : 10

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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

Sample Output

13.333
31.500

Author

CHEN, Yue

正确AC：

#include <cstdio>
#include <algorithm>
using namespace std;

struct node
{
int be;
int cat;
double bl;
}t[100000]; //数组应大点

bool cmp(const node &a,const node &b) //取地址很重要
{
if(a.bl>b.bl)
return true;
return false;
}

int main()
{
int n,m,i;
double sum;
while(~scanf("%d%d",&n,&m))
{
  if(n==-1||m==-1)      break;
  for(i=0;i<m;i++)
  {
   scanf("%d%d",&t[i].be,&t[i].cat);
   t[i].bl=(double)t[i].be/t[i].cat;
  }
  sum=0;
  sort(t,t+m,cmp);
  for(i=0;i<m;i++)
  {
   if(n-t[i].cat>0)
   {
    sum+=t[i].be;
    n-=t[i].cat;
   }
  else
   {
    sum+=n*t[i].bl;
    break;
   }
  }
  printf("%.3f\n",sum);

}
return 0;
}

或者：

#include <cstdio>
#include <algorithm>
using namespace std;

struct node
{
int be;
int cat;
double bl;
}t[100000];

bool cmp(const node &a,const node &b)
{
if(a.bl>b.bl)
return true;
return false;
}

int main()
{
int n,m,i;
double sum;
while(~scanf("%d%d",&n,&m))
{
  if(n==-1||m==-1)      break;
  for(i=0;i<m;i++)
  {
   scanf("%d%d",&t[i].be,&t[i].cat);
   t[i].bl=(double)t[i].be/t[i].cat;
  }
  sum=0;
  sort(t,t+m,cmp);
  for(i=0;i<m;i++)
  {
   if(n>0&&n>=t[i].cat)
   {
    n-=t[i].cat;
    sum+=t[i].be;
    continue;
   }
   if(n>0&&n<t[i].cat)
   {
    sum+=n*t[i].bl;
    break;
   }
  }
  printf("%.3f\n",sum);

}
return 0;
}

说明:

bool cmp(const node &a,const node &b)

运算符重载的条件是，必须存在至少一个自定义的类型。
而指针实际上就是地址，即一个32位的整数值，它们的大小是直接进行数值的比较的
如果想比较2个指针所指的对象的大小，可以对它们取 * 操作，再调用 operator < 即可。
另外，楼主写的 operator < 重载，参数应设定为 const node &, 即
bool operator < (const node &a, const node &b)
{
...
}
按LZ的方式去写的话，每次调用 < 运算符时都会存在对象的拷贝。 
再者，结构体里面的成员，在默认情况下均是 public 的权限，不需要将 bool operator < 设置为结构体的友员函数，可以直接通过对象访问成员变量。

比较:#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;
}