HDU 1009 FatMouse's Trade

最新推荐文章于 2021-05-09 21:32:11 发布

七分好月

最新推荐文章于 2021-05-09 21:32:11 发布

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本文链接：https://blog.youkuaiyun.com/qq_42506224/article/details/81265646

本题探讨了FatMouse如何利用有限的猫粮交换仓库中的JavaBeans。通过算法优化选择最佳交换比例，实现最大化的JavaBeans获取量。文章提供了一种有效的解决方案，并附带示例输入输出。

FatMouse's Trade

时间限制: 1 Sec 内存限制: 32 MB

题目描述

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.

输入

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.

输出

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.

样例输入

样例输出

2.286
2.500

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
struct node {
 
    double jz;    // jv 价值   用价值来做对比 价值高优先
    int summ;     
    int jv;
} p[1000];
 
int cmp(node a,node b) {
    if(a.jz==b.jz)
        return a.summ<b.summ;
    else return a.jz>b.jz;
}
 
int main() {
    int m,n;
    int i,j;
    int javabean,f;
    while(scanf("%d%d",&m,&n)!=EOF && m!=-1) {
        memset(p,0,sizeof(p));
        double sum=0.0;
        for(i=0; i<n; i++) {
            scanf("%d%d",&javabean,&f);
            p[i].jz=1.00*javabean/f;
            p[i].summ=f;
            p[i].jv=javabean;
        }
        sort(p,p+n,cmp);
        int max=p[0].summ;
        for(i=0; i<n; i++) {
            if(m>=max) {
                m=m-max;
                max=p[i+1].summ;
                sum=sum+p[i].jv;
            } else {
                sum=sum+m*p[i].jz;
                break;
            }
 
        }
        printf("%.3f\n",sum);
    }
    return 0;
}