SDJZU_新生_贪心_FatMouse' Trade

最新推荐文章于 2015-02-04 16:33:15 发布

原创最新推荐文章于 2015-02-04 16:33:15 发布 · 749 阅读

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#ACM #struct #程序员 #编程 #algorithm

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本文详细阐述了FatMouse在仓库中通过贪心算法最大化获取JavaBean数量的过程，包括输入输出格式、实例解析及代码实现。

昨天就是练练手，培训今天才开始

SDJZU_新生_贪心

27:25:42

30:00:00

Current Time:	2015-01-24 18:55:42	Contest Type:	Private
Start Time:	2015-01-23 15:30:00	Contest Status:	Running
End Time:	2015-01-24 21:30:00	Manager:	ACboy

	ID	Title
19 / 53	Problem A	今年暑假不AC
2 / 38	Problem B	Fence Repair
18 / 38	Problem C	FatMouse' Trade
9 / 23	Problem D	Doing Homework again
2 / 6	Problem E	Wooden Sticks
	Problem F	lines

nickname改成自己真实姓名

C - FatMouse' Trade

Crawling in process... Crawling failed Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

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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

Sample Output

       
         13.333
31.500

#include<stdio.h>
#include<algorithm>
using namespace std;
struct A
{
    int j,f;
    double z;
}a[1007];
int cmp(A a,A b)
{
    return a.z>b.z;
}
int main()
{
    int M,N;
    while(scanf("%d%d",&M,&N)!=EOF)
    {
        if(M==-1&&N==-1)
        {
            break;
        }
        A a[1007];
        int i;
        for(i=0;i<N;i++)
        {
            scanf("%d%d",&a[i].j,&a[i].f);
            a[i].z=((double)a[i].j/a[i].f);
        }
        sort(a,a+N,cmp);
        double b=0;
        for(i=0;M>0&&i<N;i++)
        {
            if(M>=a[i].f)
            {
                M-=a[i].f;
                b+=a[i].j;
            }
            else
            {
                b+=(((double)M/a[i].f)*a[i].j);
                M=0;
            }
        }
        printf("%.3f\n",b);
    }
    return 0;
}

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