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C - FatMouse' Trade

2024-08-08 08:04:05   222人阅读

FatMouse‘ Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 45965    Accepted Submission(s): 15395


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
 
 
首先这是一道贪心算法，当f/j越大时代表这样越有利，所以进行排序；
但也有值得注意的几组数据（来自杭电的大神）
此题除了要满足例子以外，还要满足一些条件才能真正算ac：0 11 01.0001 00.0005 410000 52000 2100 0300 010400.000数据类型用double,就这样
 1 #include<stdio.h> 2 int j[1010],f[1010]; 3 double a[1010]; 4 void paixu(double a[],int k[],int f[],int n) 5 { 6     int i=0,j=n-1,h=k[0],m=f[0]; 7     double  t=a[0]; 8     if(n>1){ 9         while(i<j)10         {11             for(;i<j;j--)12             if(a[j]>t){13                 a[i]=a[j];14                 f[i]=f[j];15                 k[i]=k[j];16                 i++;17                 break;18             }19             for(;i<j;i++)20             {21                 if(a[i]<t){22                     a[j]=a[i];23                     f[j]=f[i];24                     k[j]=k[i];25                     j--;26                     break;27                 }28             }29         }30         a[i]=t;31         f[i]=m;32         k[i]=h;33         paixu(a,k,f,i);34         paixu(a+i+1,k+i+1,f+i+1,n-i-1);35     }36 }37 void hanshu(int m,int n)38 {39     int i;40     double  sum=0;41     for(i=0;i<n;i++)42     {43         if(m>=f[i]){44             sum=sum+j[i];45             m=m-f[i];46         }47         else{48             sum=sum+1.0*m/f[i]*j[i];49             break;50         }51     }52     printf("%0.3f\n",sum);53 }54 int main()55 {56     int n,m,i,t;57     while(1)58     {scanf("%d%d",&m,&n);59     if(m==-1&&n==-1)break;60     for(i=0;i<n;i++)61         {scanf("%d%d",&j[i],&f[i]);62         if(f[i]!=0)a[i]=1.0*j[i]/f[i];63         else a[i]=1020;64         }65         paixu(a,j,f,n);66         hanshu(m,n);67         }68     return 0;69 70 }

 

 

C - FatMouse' Trade


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