hdu 1009:FatMouse' Trade(贪心)
2014-01-01 13:35
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FatMouse' Trade
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 36632 Accepted Submission(s): 12064
[align=left]Problem Description[/align]
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.
[align=left]Input[/align]
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.
[align=left]Output[/align]
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.
[align=left]Sample Input[/align]
5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1
[align=left]Sample Output[/align]
13.333
31.500
[align=left]Author[/align]
CHEN, Yue
[align=left]Source[/align]
ZJCPC2004
[align=left]Recommend[/align]
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水题,读不懂题意是个坑,英语是个大问题啊……
#include <iostream> #include <iomanip> using namespace std; int main() { int N,M; while(cin>>M>>N,M!=-1 || N!=-1){ int J[1001],F[1001]; for(int i=1;i<=N;i++){ cin>>J[i]>>F[i]; } //ÅÅÐò for(int i=1;i<=N-1;i++) for(int j=1;j<=N-i;j++){ if((double)J[j]/F[j]<(double)J[j+1]/F[j+1]){ int t; t=F[j];F[j]=F[j+1];F[j+1]=t; t=J[j];J[j]=J[j+1];J[j+1]=t; } } int fs=1,fe=1; //¼ÆËã double res=0; for(int i=1;i<=N;i++){ if(M-F[i]>=0){ res+=J[i]; M-=F[i]; } else{ res+=M/(double)F[i]*(double)J[i]; break; } } cout<<setiosflags(ios::fixed); cout<<setprecision(3); cout<<res<<endl; } return 0; }
Freecode : www.cnblogs.com/yym2013
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