HDU 1009 FatMouse' Trade (贪心算法)

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FatMouse' Trade Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 67076    Accepted Submission(s): 22818 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   #include #include #include #include using namespace std; struct node { double wei; double cost; double average; }a[10001]; bool cmp(node a,node b) { return a.average>b.average; } int main() { double m; int n; while(cin>>m>>n) { if(m==-1&&n==-1) break; for(int i=1;i<=n;i++) { cin>>a[i].wei>>a[i].cost; a[i].average=a[i].wei/a[i].cost; } sort(a+1,a+n+1,cmp); double cnt=0; for(int i=1;i<=n;i++) { if(m>=a[i].cost) { cnt+=a[i].wei; m-=a[i].cost; } else { cnt+=m*(a[i].average); m=0; } if(m==0) break; } printf("%.3lf\n",cnt); } return 0; }