FatMouse' Trade
来源:互联网 发布:mac本怎么退出全屏显示 编辑:程序博客网 时间:2024/06/11 04:33
A - FatMouse' Trade
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64uDescription
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#include<stdio.h>#include<string.h>int main(){ int m,n,j[1005],f[1005],z,t=0; double e[1005]; scanf("%d%d",&m,&n); while(m!=-1&&n!=-1) { t++; z=0; for(int x=0;x<n;x++) scanf("%d%d",&j[x],&f[x]); double a[1005],s=0,c; int k; for (int i=0;i<n;i++) a[i]=(double)j[i]/f[i]; for(int i=0;i<n-1;i++) for(int w=i+1;w<n;w++) if(a[i]<a[w]) { c=a[i];a[i]=a[w];a[w]=c; k=j[i];j[i]=j[w];j[w]=k; k=f[i];f[i]=f[w];f[w]=k; } for(int i=0;i<n;i++) if(f[i]<=m) { s=s+j[i];m=m-f[i];if(m==0) break;} else {s=s+(double)m/f[i]*j[i];break;} e[t-1]=s; memset(j,0,sizeof(j)); memset(f,0,sizeof(f)); scanf("%d%d",&m,&n) ; } for(int d=0;d<t;d++) printf("%.3f\n",e[d]); return 0;}
0 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
- ASIDownload下载显示的用法
- hibernate连接Oracle数据库出现关闭的连接问题
- Leetcode Surrounded Regions 解题报告
- HDU 2196
- android实现数据库和UI同步更新
- FatMouse' Trade
- 音频采样位数,采样率,比特率
- SSH Secure Shell Client用密钥认证登录linux服务器
- css样式大全
- 《学习OpenCV》练习7-6
- 常用的正则表达式
- myeclipse8.5 启动时不弹出工作空间的选择
- 我的苹果开发学习笔记
- NVelocity入门,为服务器与客户端传输xml数据,实现Ajax通信铺平道路