FatMouse' Trade
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 49098 Accepted Submission(s): 16485
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.
简单贪心....
需要注意的是数据是非负,所以有0的情况要考虑周全,基本都要特殊处理。
多WA了三次,不知道为什么交C++可以过,交G++就不行。
#include <iostream> #include <cstring> #include <algorithm> #include <cstdio> #include <iomanip> using namespace std; int main() { int m,n,f[1500],j[1500]; double scale[1500]; double ans,sum; while(scanf("%d %d",&m,&n)!=EOF&&(m!=-1)) { ans=0; // if (m==0) memset(scale,0,sizeof(scale)); memset(f,0,sizeof(f)); memset(j,0,sizeof(j)); // bool flag=false; for (int i=1;i<=n;i++) { scanf("%d %d",&j[i],&f[i]); if (f[i]!=0) scale[i]=(double)j[i]*1.0/f[i]; else if (j[i]!=0) { ans=ans+j[i]; } } // if (flag) {cout<<fixed<<setprecision(3)<<ans<<endl;continue;} for (int i=1;i<=n-1;i++) for (int k=i+1;k<=n;k++) if (scale[i]<scale[k]) { swap(scale[i],scale[k]); swap(j[i],j[k]); swap(f[i],f[k]); } sum=0; int i=1; while (m>=sum&&i<=n) { ans=ans+j[i]; sum=sum+f[i]; i++; } i--; ans=ans-j[i]; sum=sum-f[i]; ans=ans+(m-sum)*scale[i]; cout<<fixed<<setprecision(3)<<ans<<endl; } return 0; }