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
题解：贪心求解，每一次选择性价比最高的那一个；
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; struct app { double a; double b; double p; }; bool cmp(struct app a,struct app b) { return a.p>b.p; } int main() { int p,i,j,m,u; double a,k,pp; struct app b[1111]; for(;scanf("%d%d",&p,&m)!=-1;) { if(p==-1&&m==-1) break; for(i=0;i<m;i++) { scanf("%lf%lf",&b[i].a,&b[i].b); b[i].p=b[i].a/b[i].b; } sort(b,b+m,cmp); u=0;pp=0.0; for(i=0;i<m;i++) { pp+=b[i].a; u+=b[i].b; if(u>p) { pp=(pp-b[i].a)+(p-u+b[i].b)*b[i].p; break; } } printf("%.3lf\n",pp); } return 0; }