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贪心专题 HDU

2017-05-16 12:55 288 查看


FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 77018    Accepted Submission(s): 26456


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

 

Recommend

JGShining

 
题意,这里有一个奇葩的老鼠和一群奇葩的猫,老鼠喜欢吃javabean,但是他手里只有猫咪爱 吃的食物。每一个猫咪看守一部分javabean,允许老鼠拿他爱吃的食物来换,多少食物换多javabean,不同的猫咪要求的食物及提供的javabean不一样,让你算算老鼠最大可以换多少javabean。
贪心题,老鼠肯定是先换取最实惠的,他那么聪明
ac代码(提交使用G++的时候,要记得double输出使用%f)

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>

using namespace std;

struct node{
int jav,foo;
double ave;
}nn[1005];

bool cmp(node x,node y){
return x.ave > y.ave;
}
int m,n;

int main()
{

while(~scanf("%d%d",&m,&n),(m!=-1 || n!=-1)){

for(int i=1;i<=n;i++){
scanf("%d%d",&nn[i].jav,&nn[i].foo);
nn[i].ave=nn[i].jav*1.0/nn[i].foo;
}
sort(nn+1,nn+1+n,cmp);
double ans=0;
for(int i=1;i<=n;i++){
if(m > nn[i].foo){
ans+=nn[i].jav;
m=m-nn[i].foo;
}
else if(m > 0){
ans+=nn[i].ave*m;
break;
}
else
break;
}

printf("%.3f\n",ans);
}

return 0;
}
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