FatMouse’ Trade Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 119147 Accepted Submission(s): 41165 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**
思路：非常简单的贪心题型，优先选择性价比高的房间即可。 代码：
#include <iostream> using namespace std; #define MAXMUM 1005 int J[MAXMUM]; int F[MAXMUM]; double value[MAXMUM]; int main(){ int M,N; while(cin>>M>>N&&M!=-1&&N!=-1){ for(int i=0;i<N;i++){ cin>>J[i]>>F[i]; value[i] = 1.0*J[i]/F[i]; } for(int i=0;i<N;i++){ for(int j=i;j<N;j++){ if(value[i]<value[j]){ swap(value[i],value[j]); swap(J[i],J[j]); swap(F[i],F[j]); } } } int now = 0; double num = 0; while(M&&now<N){ if(M>=F[now]){ M-=F[now]; num+=J[now]; } else{ num+=M*value[now]; M = 0; } now++; } printf("%.3f\n",num); } return 0; }