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A - FatMouse' Trade
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 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<stdio.h> #include<algorithm> using namespace std; struct p { double a; double b; double c; }; bool compare(p A,p B) { return A.c>B.c; } int main() { int m,n,i,j; while(scanf("%d%d",&m,&n)!=EOF) { if(m==-1&&n==-1) break; p t[1001]; for(i=0;i<n;i++) { cin>>t[i].a>>t[i].b; t[i].c=t[i].a/t[i].b; } sort(t,t+n,compare); double count=0; for(i=0;i<n;i++) { count+=t[i].b; if(count>m) break; } double sum=0; //if(i==0) //printf("%.3lf\n",t[0].c*m); if(count<m) { for(j=0;j<n;j++) sum+=t[j].a; printf("%.3lf\n",sum); } else { for(j=0;j<i;j++) { sum+=t[j].a; m-=t[j].b; } printf("%.3lf\n",sum+m*t[i].c); } } return 0; }
#include<iostream> #include<stdio.h> using namespace std; int main() { int m,n; int i,j; double a[1002],b[1002]; double c[1002]; while(scanf("%d%d",&m,&n)!=EOF) { if(m==-1&&n==-1) break; for(i=0;i<n;i++) { cin>>a[i]>>b[i]; c[i]=a[i]/b[i]; } for(i=0;i<n;i++) { for(j=0;j<n-i-1;j++) { if(c[j+1]>c[j]) { swap(b[j],b[j+1]); swap(a[j],a[j+1]); swap(c[j],c[j+1]); } } } double count=0; for(i=0;i<n;i++) { count+=b[i]; if(count>m) break; } double sum=0; if(count<m) { for(j=0;j<n;j++) sum+=a[j]; printf("%.3lf\n",sum); } else { for(j=0;j<i;j++) { sum+=a[j]; m-=b[j]; } printf("%.3lf\n",sum+m*c[i]); } } return 0; }
A - FatMouse' Trade
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