HDU 1009:FatMouse' Trade

HDU 1009:FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 60163    Accepted Submission(s): 20217

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

Source

​​ZJCPC2004​​

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#include #include using namespace std;struct Node{ double j,f,p;} node[10000];int cmp(Node x,Node y){ return x.p>y.p;}int main(){ int m,n; while(~scanf("%d%d",&n,&m) && (m!=-1 || n!=-1)) { double sum = 0; int i; for(i = 0; inode[i].f) { sum+=node[i].j; n-=node[i].f; } else { sum+=node[i].p*n; break; } } printf("%.3lf\n",sum); } return 0;}

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