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.
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.
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.
4 2 4 7 1 3 5 5 4 8 3 8 1 2 2 5 2 4 -1 -1
2.286 2.500
//思想 按照“单位价格”进行从小到大排序 剩下的类似背包看是否能装进去！（此题的价格也写进了结构体 对单价进行排序也需要对相应的 mouse cat 进行排序）
#include<stdio.h> #include<algorithm> using namespace std; struct Node { double mouse,cat,aver;//是double 不是int 三个！！！ } str[1005]; bool cmp(Node x,Node y) { return x.aver<y.aver;//按照价格从小到大排序 } int main() { int M,N; while(~scanf("%d %d",&M,&N)&&(M!=-1)&&(N!=-1)) { for(int i=0; i<N; i++) { scanf("%lf %lf",&str[i].mouse,&str[i].cat); str[i].aver=str[i].cat/str[i].mouse; } sort(str,str+N,cmp); // for(int i=0;i<N;i++){ // printf("%lf %lf %lf\n",str[i].mouse,str[i].cat,str[i].aver); // } double sum=0.0; for(int i=0; i<N; i++) { if(str[i].cat<M) {//类似于背包 M-=str[i].cat; sum+=str[i].mouse; } else { sum+=str[i].mouse*M/str[i].cat; break;//别忘了！ } } printf("%.3lf\n",sum); } return 0; }