hdu1009 FatMouse' Trade(贪心)
2015-05-20 14:53
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FatMouse' Trade
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 51313 Accepted Submission(s): 17182
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.
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
Author
CHEN, Yue
Source
ZJCPC2004
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解析:贪心。
按照每个房间的 j[i]/f[i] 的值,有大到小进行排序,从最大的开始,能卖多少就买多少,直到买完为止。
需要注意的是,有的房间的 f[i] 为 0 的情况,就是不要猫粮,直接白送(我靠,真是一只好猫,跟老鼠做交易就算了,还白送,我要是主人的话,我保证绝对不打死它)。
代码:
#include<cstdio> #include<algorithm> #define maxn 1000 using namespace std; struct tnode{int x,y;double z;}; tnode a[maxn+20]; bool cmp(tnode p,tnode q) { return p.z>q.z; } int main() { freopen("1.in","r",stdin); int m,n,i,j,k,sum; double ans; while(scanf("%d%d",&m,&n)&&m!=-1) { ans=0,sum=0; for(i=1;i<=n;i++) { scanf("%d%d",&j,&k); if(k==0){ans+=j;continue;} a[++sum].x=j,a[sum].y=k; a[sum].z=j*1.0/k; } sort(a+1,a+sum+1,cmp); for(i=1;m>=0 && i<=sum;i++) { k=min(m,a[i].y); ans+=a[i].z*k; m-=k; } printf("%.3lf\n",ans); } return 0; }
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