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  • FatMouse' Trade

    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
     
     1 #include <stdio.h> 
     2 #define N 1001
     3 
     4 int main(){
     5     int m;
     6     int n;
     7     int i;
     8     int j;
     9     int J[N];
    10     int F[N];
    11     double unit[N];
    12     double temp;
    13     double get;
    14 
    15     while(1){
    16         scanf("%d%d",&m,&n);
    17 
    18         if(m==-1 && n==-1)
    19             break;
    20 
    21         get=0;
    22         for(i=0;i<n;i++){
    23             scanf("%d%d",&J[i],&F[i]);
    24             unit[i]=(double)J[i]/F[i];
    25         }
    26 
    27         for(i=0;i<n-1;i++){
    28             for(j=i+1;j<n;j++){
    29                 if(unit[i]<unit[j]){
    30                     temp=unit[i];
    31                     unit[i]=unit[j];
    32                     unit[j]=temp;
    33 
    34                     temp=J[i];
    35                     J[i]=J[j];
    36                     J[j]=temp;
    37 
    38                     temp=F[i];
    39                     F[i]=F[j];
    40                     F[j]=temp;
    41                 }
    42             }
    43         }
    44 
    45         
    46         for(i=0;i<n;i++){
    47             if(m>=F[i]){
    48                 m=m-F[i];
    49                 get+=J[i];   //这里J[i]改成unit[i]*F[i]会丢失精度导致出错
    50 
    51                 if(m==0)
    52                     break;
    53             }
    54 
    55             else{
    56                 get+=unit[i]*m;
    57                     
    58                 break;
    59             }
    60         }
    61         printf("%.3lf
    ",get);
    62     }    
    63     return 0;
    64 }
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  • 原文地址:https://www.cnblogs.com/zqxLonely/p/4064001.html
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