FatMouse' Trade
来源:互联网 发布:人工智能的基础 编辑:程序博客网 时间:2024/06/10 16:36
FatMouse' Trade
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 16 Accepted Submission(s) : 9
Font: Times New Roman | Verdana | Georgia
Font Size: ← →
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.
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 37 24 35 220 325 1824 1515 10-1 -1
Sample Output
13.33331.500
Author
Source
ZJCPC2004
#include<stdio.h>struct node{ int a; int b; double c;};struct node st[1000],t;int main(){ int n,m,i,j,s1,flag; double sum; while(scanf("%d%d",&n,&m)!=EOF) { if(n==-1&&m==-1) break; for(i=0;i<=m-1;i++) { scanf("%d%d",&st[i].a,&st[i].b); st[i].c=(double)st[i].a/st[i].b; } for(i=0;i<=m-2;i++) for(j=i+1;j<=m-1;j++) { if(st[i].c<st[j].c) { t=st[i]; st[i]=st[j]; st[j]=t; } } s1=0; sum=0; flag=1; for(i=0;i<=m-1&&flag==1;i++) { s1+=st[i].b; if(s1<=n) { sum+=(double)st[i].a; } else { s1=n-(s1-st[i].b); sum+=(double)s1*st[i].c; flag=0; } } printf("%.3lf\n",sum); } return 0;}
- FatMouse' Trade
- FatMouse' Trade
- FatMouse' Trade
- FatMouse' Trade
- FatMouse' Trade
- FatMouse' Trade
- FatMouse' Trade
- FatMouse' Trade
- FatMouse' Trade
- FatMouse' Trade
- FatMouse' Trade
- FatMouse' Trade
- FatMouse' Trade
- FatMouse' Trade
- FatMouse' Trade
- FatMouse' Trade
- FatMouse' Trade
- FatMouse' Trade
- 网站发布的一点小问题
- ThreadLocal类讲解及实例
- IT人应该学的厚黑学
- JAVA螺旋队列
- jQuery小试之用户名校验
- FatMouse' Trade
- hive指南
- android <shape>各个属性
- effective java(方法)
- 编码规范
- 高级线程开发(线程池、资源封锁和队列)
- 排序
- informix
- Number Sequence