题目1433:FatMouse
题目1433:FatMouse
时间限制:1 秒
内存限制:128 兆
特殊判题:否
- 题目描述:
-
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.
- 输入:
-
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.
- 输出:
-
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.
- 样例输入:
-
5 3 7 2 4 3 5 2 20 3 25 18 24 15 15 10 -1 -1
- 样例输出:
-
13.333 31.500
#include <iostream> #include<stdio.h> #include<algorithm> using namespace std; struct goods { double j; double f; double s; bool operator <(const goods & A) const { return s>A.s; } } buf[1000]; int main() { double m; int n; while(scanf("%lf%d",&m,&n)!=EOF) { if(m==-1&&n==-1) break; for(int i=0; i<n; i++) { scanf("%lf %lf",&buf[i].j,&buf[i].f); buf[i].s=buf[i].j/buf[i].f; } sort(buf,buf+n);//按照性价比降序排序 int idx=0;//当前货物下标 double ans=0;//累加所得货物总重量 while(idx<n&&m>0)//循环条件为货物还有剩余(idx<n),钱还有剩余(m>0) { if(m>buf[idx].f) { ans+=buf[idx].j; m-=buf[idx].f; } else { ans+=m*buf[idx].s; m=0; } idx++; } printf("%.3lf\n",ans); } return 0; } /************************************************************** Problem: 1433 User: zhuoyuezai Language: C++ Result: Accepted Time:10 ms Memory:1544 kb ****************************************************************/