1660 - FatMouse's Trade
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.
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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.
Examples
Input
4 2 4 7 1 3 5 5 4 8 3 8 1 2 2 5 2 4 -1 -1
Output
2.286 2.500
Solution C
#include <stdio.h> int main() { int m,n,i=0,t=0; double j[1000],f[1000],sum=0; while(scanf("%d %d",&m,&n)!=EOF && m+n>0) { for(i=0;i<n;i++) { scanf("%lf %lf",&j[i],&f[i]); } for(i=0;i<n;i++) { for(t=0;t<n;t++) { if(j[i]/f[i]>j[t]/f[t]) { double temp; temp=j[i]; j[i]=j[t]; j[t]=temp; temp=f[i]; f[i]=f[t]; f[t]=temp; } } } for(i=0;i<n;i++) { if(m<f[i]) { sum+=m*j[i]/f[i]; break; } else { sum+=j[i]; m-=f[i]; } } printf("%.3f\n",sum); sum=0; } return 0; }
Solution C++
//注意m和f可以为0 #include <iostream> #include <algorithm> #include <cstdio> #include <string.h> #include <queue> #include <stdlib.h> using namespace std; typedef struct node { int j; int f; double d; }MC; MC a[1003]; bool cmp(MC a,MC b) { return a.d>b.d; } int main() { int n,m,i; double t; while(scanf("%d%d",&n,&m)) { if(n==-1&&m==-1) break; for(i=0;i<m;i++) { scanf("%d%d",&a[i].j,&a[i].f); a[i].d=1.0*a[i].j/a[i].f; } sort(a,a+m,cmp); t=0; for(i=0;i<m;i++) { if(n>a[i].f) { t+=a[i].j; n-=a[i].f; } else { t+=n*a[i].d; break; } } printf("%.3lf\n",t); } return 0; }