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
Author
CHEN, Yue
Source
Recommend
JGShining
(转自hdu)
按照f / j排序
Code
1 /** 2 * acm.hdu.edu.cn 3 * Problem#1009 4 * Accepted 5 * Time:46ms 6 * Memory:1596k 7 */ 8 #include<iostream> 9 #include<cstdio> 10 #include<cctype> 11 #include<cstring> 12 #include<cstdlib> 13 #include<cmath> 14 #include<fstream> 15 #include<sstream> 16 #include<algorithm> 17 #include<map> 18 #include<set> 19 #include<queue> 20 #include<vector> 21 #include<stack> 22 using namespace std; 23 typedef bool boolean; 24 #define INF 0xfffffff 25 #define smin(a, b) a = min(a, b) 26 #define smax(a, b) a = max(a, b) 27 template<typename T> 28 inline void readInteger(T& u){ 29 char x; 30 int aFlag = 1; 31 while(!isdigit((x = getchar())) && x != '-'); 32 if(x == '-'){ 33 x = getchar(); 34 aFlag = -1; 35 } 36 for(u = x - '0'; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - '0'); 37 ungetc(x, stdin); 38 u *= aFlag; 39 } 40 41 typedef class Data{ 42 public: 43 int f; 44 int j; 45 double b; 46 }Data; 47 48 int n, m; 49 Data* datas; 50 51 inline boolean init() { 52 readInteger(n); 53 readInteger(m); 54 if(n == -1 && m == -1) return false; 55 datas = new Data[(const int)(m + 1)]; 56 for(int i = 1; i <= m; i++){ 57 readInteger(datas[i].j); 58 readInteger(datas[i].f); 59 datas[i].b = datas[i].j * 1.0 / datas[i].f; 60 } 61 return true; 62 } 63 64 boolean cmpare(const Data& a, const Data& b){ 65 return a.b > b.b; 66 } 67 68 double result; 69 inline void solve() { 70 result = 0.0; 71 sort(datas + 1, datas + m + 1, cmpare); 72 int i = 1; 73 while(n >= 0 && i <= m){ 74 int buy = min(n, datas[i].f); 75 if(buy == datas[i].f) result += datas[i].j; 76 else result += buy * 1.0 / datas[i].f * datas[i].j; 77 n -= buy; 78 i++; 79 } 80 printf("%.3lf ", result); 81 } 82 83 int main(){ 84 while(init()){ 85 solve(); 86 delete[] datas; 87 } 88 return 0; 89 }