• # K贪心

```<span style="color:#330099;">/*
K - 贪心 基础
Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u
Submit

Status
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
BY Grant Yuan
2014.7.14
贪心
*/
/*
描写叙述
有一个投资人。他有金钱m，可选投资项目n个。对于每一个投资项目，投入金钱c则获得收益f。要求对每一个项目所投入的资金r应该在0到c之间（即0<=r<=c），获得的收益为r*f/c。如今你的任务就是求出投资人能获得的最大收益。

输入
包含多个測例。每一个測例第一行为两个整数m,n，分别表示金钱数和项目个数。接下来n行，每行两个整数为收益f和投入c。输入最后以两个-1结尾。

输出
包含n行，每行为相应測例的最大收益。保留三位小数printf("%.3lf
",result);。

输入例子
3 3
4 2
2 1
3 1
4 3
4 2
5 3
4 3
-1 -1

输出例子
7.000
7.333
By yuan.c
20146/22

提示
*/
#include<stdio.h>
#include<iostream>
#include<cstring>
#include<cstdlib>
using namespace std;
int N;
int M;

double p[1002];
int m[1002];
int s[1002];
double sum=0;

void sort()
{int t;double l;
for(int i=0;i<N-1;i++)
for(int j=i;j<N;j++)
{
if(p[i]<p[j]){
l=p[i],p[i]=p[j],p[j]=l;
t=m[i],m[i]=m[j],m[j]=t;
t=s[i],s[i]=s[j],s[j]=t;
}
}
}

int main()
{
while(1){
scanf("%d %d",&M,&N);
if(N==-1&&M==-1)
break;
for(int i=0;i<N;i++)
scanf("%d %d",&s[i],&m[i]);
for(int i=0;i<N;i++)
p[i]=s[i]*1.0/m[i]*1.0;
sort();
for(int i=0;i<N;i++)
if(M>=m[i])
sum+=s[i],M-=m[i];
else {sum+=M*p[i];
break;
}
printf("%.3lf
",sum);
sum=0;
memset(s,0,100);
memset(m,0,100);
memset(p,0,100);
}
return 0;
}
</span>```

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• 原文地址：https://www.cnblogs.com/lcchuguo/p/5236075.html