FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 67262    Accepted Submission(s): 22899 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

典型的贪心算法

用sort排序一下就好了；

#include <stdio.h> #include <iostream> #include <algorithm> using namespace std; struct asd{ int j; int f; double ave; }mou[3005]; bool cmp(asd a,asd b) { return a.ave>b.ave; } int main() { int m,n; while(~scanf("%d%d",&m,&n)&&(m!=-1||n!=-1)) { int i; for(i=0;i<n;i++) { scanf("%d%d",&mou[i].j,&mou[i].f); mou[i].ave=(1.0*mou[i].j)/(1.0*mou[i].f); } sort(mou,mou+n,cmp); double sum=0; for(i=0;i<n;i++) { if(m>mou[i].f) { sum+=mou[i].j; m=m-mou[i].f; } else { sum+=mou[i].ave*m; m=0; break; } } printf("%.3lf\n",sum); } return 0; }