HDOJ1009

 1 #include <iostream>
 2 #include <iomanip>
 3 #include <string.h>
 4 
 5 using namespace std;
 6 
 7 double J[1000];
 8 double F[1000]; 
 9 double Ratio[1000];
10 
11 int FindMax()   //返回最大值的下标
12 {
13     double max = 0;
14     int key;
15     for(int k = 0;k < 1000;k ++)
16         if(max < Ratio[k])
17         {
18             max = Ratio[k];
19             key = k;
20         }
21     return key;
22 }
23 
24 int main()
25 {
26 
27     int m,n;
28     int i,j,key;
29     double result;
30     while((cin>>m>>n))
31     {                    //最外层循环，表示测试用例的个数
32         if((m == -1) && (n == -1)) break;
33         memset(J,0,1000 * sizeof(double));
34         memset(F,0,1000 * sizeof(double));
35         result = 0;
36         for(i = 0;i < n;i++)
37            cin>>J[i]>>F[i];   //输入
38         for(i = 0;i < n;i++)
39         {
40             if(F[i] != 0)
41                 Ratio[i] = J[i]/F[i]; //计算单位的比例
42             else
43                 Ratio[i] = J[i];
44         }
45 
46 
47         //特殊情况，当F的值为0时,首先全部累加到result
48         if(n == 0)
49         {
50             result == 0;
51             cout<<setiosflags(ios::fixed)<< setprecision(3)<<result<<endl;
52             continue;
53         }
54         for(j = 0;j < n;j ++)
55         {
56             if(F[j] == 0)
57                result += J[j];
58         }
59         if(m == 0)
60             cout<<setiosflags(ios::fixed)<< setprecision(3)<<result<<endl;
61 
62         //计算一个测试用例的结果
63         key = FindMax();
64         while(m > 0)
65         {
66             if(m <= F[key])
67             {
68                 result += Ratio[key] * m;
69                 cout<<setiosflags(ios::fixed)<< setprecision(3)<<result<<endl;
70                 m = 0;  //准备退出，计算完毕
71             }
72             else
73             {
74                 result += Ratio[key] * F[key];
75                 m = m - (int)F[key];  //m减少
76                 Ratio[key] = 0.0;  //更新比例的值
77                 key = FindMax();  //更新key的值
78             }
79         }
80 
81     }
82 
83     return 0;
84 }