HDUOJ1009 FatMouse’ Trade

 

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15808    Accepted Submission(s): 4746

Problem DescriptionFatMouse 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. 

 

InputThe 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. 

 

OutputFor 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 37 24 35 220 325 1824 1515 10-1 -1

 

 

Sample Output

13.33331.500

 

 

AuthorCHEN, Yue#include<iostream><br />#include<algorithm><br />using namespace std;<br />struct fat { //定義結構體<br /> int x, y;<br /> double k;<br />} p[1001];<br />bool cmp(fat a, fat b) {<br /> return a.k > b.k;<br />}<br />int main() {<br /> int i, j, m, n;<br /> while (cin >> m >> n, m + n >= 0) {<br /> for (i = 0; i < n; i++) {<br /> cin >> p[i].x >> p[i].y;<br /> p[i].k = p[i].x * 1.0 / p[i].y;<br /> }<br /> sort(p, p + n, cmp); // 排序<br /> double sum = 0, tot = 0;<br /> for (i = 0; i < n; i++) {<br /> sum += p[i].x;<br /> tot += p[i].y;<br /> if (tot == m)<br /> break;<br /> if (tot > m) {<br /> tot -= p[i].y;<br /> sum -= p[i].x;<br /> sum += (((m - tot) * 1.0 / p[i].y) * p[i].x);<br /> break;<br /> }<br /> }<br /> printf("%.3lf/n", sum);<br /> }<br /> return 0;<br />}