HDU 1009 FatMouse' Trade(贪心)
2012-03-13 23:41
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FatMouse' Trade
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 20968 Accepted Submission(s): 6501
[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]
JGShining
很简单的贪心题,
先按照价值/代价的比值来排序,肯定是先买比值大的。
代码如下:
#include<stdio.h> #include<stdlib.h> const int MAXN = 1010; struct node { double j,f; double r; }a[MAXN]; int cmp(const void *a,const void *b)//从大到小排序 { struct node *c=(node *)a; struct node *d=(node *)b; if(c->r > d->r) return -1; else return 1; } int main() { int N; double M; double ans; while(scanf("%lf%d",&M,&N)) { if(M==-1&&N==-1) break; for(int i=0;i<N;i++) { scanf("%lf%lf",&a[i].j,&a[i].f); a[i].r=(double)a[i].j/a[i].f; } qsort(a,N,sizeof(a[0]),cmp); ans=0; for(int i=0;i<N;i++) { if(M>=a[i].f) { ans+=a[i].j; M-=a[i].f; } else { ans+=(a[i].j/a[i].f)*M; break; } } printf("%.3lf\n",ans); } return 0; }
sort排序版本:
#include<stdio.h> #include<stdlib.h> #include<algorithm> using namespace std; const int MAXN = 1010; struct node { double j,f; double r; }a[MAXN]; /* int cmp(const void *a,const void *b)//从大到小排序 { struct node *c=(node *)a; struct node *d=(node *)b; if(c->r > d->r) return -1; else return 1; } */ bool cmp(node a,node b) { return a.r > b.r; } int main() { int N; double M; double ans; while(scanf("%lf%d",&M,&N)) { if(M==-1&&N==-1) break; for(int i=0;i<N;i++) { scanf("%lf%lf",&a[i].j,&a[i].f); a[i].r=(double)a[i].j/a[i].f; } //qsort(a,N,sizeof(a[0]),cmp); sort(a,a+N,cmp); ans=0; for(int i=0;i<N;i++) { if(M>=a[i].f) { ans+=a[i].j; M-=a[i].f; } else { ans+=(a[i].j/a[i].f)*M; break; } } printf("%.3lf\n",ans); } return 0; }
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