FatMouse' Trade

发布时间:2014-10-22 18:28:29编辑:www.fx114.net 分享查询网我要评论
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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.    Sample Input 5 3 7 2 4 3 5 2 20 3 25 18 24 15 15 10 -1 -1   Sample Output 13.333 31.500 #include<iostream> #include<string.h> #include<stdio.h> #include<ctype.h> #include<algorithm> #include<stack> #include<queue> #include<set> #include<math.h> #include<vector> #include<deque> #include<list> using namespace std; struct wp { double d,f,x;//float型数据在相除的时候会有偏差,用double型比较好,减少误差 }p[1010]; int cmp(wp a,wp b) { return a.x>b.x; } int main() { int m,n,i; double sum=0; while((scanf("%d%d",&m,&n)!=EOF) &&(n!=-1 || m!=-1)) { sum=0; for(i=0;i<n;i++) { cin>>p[i].d>>p[i].f; p[i].x=p[i].d/p[i].f;//求性价比 } sort(p,p+n,cmp);//根据性价比从大到小排序 for(i=0;i<n;i++) { if(p[i].f<m)//如果占用空间比总空间小,直接加上 { sum+=p[i].d; m-=p[i].f; } else//否则用性价比的(相当于数学中的权)去乘以剩下的空间 { sum+=p[i].x*m; break; } } printf("%.3f\n",sum); } return 0; }

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