FatMouse' Trade
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 92544 Accepted Submission(s): 32114
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
解题思路:贪心,开一个结构体后按照平均值从大到小排序,先换掉平均值大的房间里的JavaBean,输出用printf输出,用setpricision会wa,原因在于setprecision会四舍五入丢失精度。
#include#include using namespace std;struct jb{ double j,f,avg;};bool cmp(jb x,jb y){ return x.avg >n>>m&&n!=-1) { jb s[1001]; for(int i=0;i >s[i].j>>s[i].f;s[i].avg=s[i].j/s[i].f;} sort(s,s+m,cmp); double sum=0; for(int i=m-1;i>=0;--i) { if(n>=s[i].f) { n-=s[i].f; sum+=s[i].j; } else if(n 0) { sum+=s[i].avg*n; break; } } printf("%.3lf\n",sum); //cout<< < <