hdoj1009 贪心

原文：
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

这是我的第一篇博客，如有不好的地方，请多多指教。

问题描述：
给你m pounds的猫粮，你有n种兑换为JavaBean的方式。接下来n行是兑换的方式，比如：第一次给你5 pounds的猫粮，有3种兑换方式：2 pounds 猫粮兑换7 pounds JavaBean，3 pounds 猫粮兑换4 pounds JavaBean，2 pounds 猫粮兑换5 pounds JavaBean。要求这5 pounds 猫粮最多能够兑换的JavaBean的数量。

思路：
要求最多能兑换多少JavaBean，则可以用贪心算法：求出兑换比例 JavaBean / 猫粮（题中为 J[i] / F[i]），找到最大的兑换比例并优先考虑。最后剩余不够整体兑换的猫粮就用相应的比例计算即可。

代码如下：

#include <iostream>#include <cstdio>using namespace std;typedef struct gui{    int j;    int f;    double bili;}Gui;Gui javabin[1005];void qsort(int left,int right,Gui javabin[]){    if (left >= right){        return;    }    Gui key = javabin[left];    int i = left,j1 = right;    while (i < j1){        while (i < j1 && key.bili <= javabin[j1].bili){            j1--;        }        javabin[i] = javabin[j1];        while (i < j1 && key.bili >= javabin[i].bili){            i++;        }        javabin[j1] = javabin[i];        }    javabin[i] = key;    qsort(left,i-1,javabin);    qsort(i+1,right,javabin);}void test(){    int m,n;    while (cin >> m >> n){        if (m == -1 && n == -1){            break;        }        for (int i = 0;i < n;i++){            cin >> javabin[i].j >> javabin[i].f;            if (javabin[i].f == 0){                javabin[i].bili = (double)(javabin[i].j * 10000);            }            else{                javabin[i].bili = (double)javabin[i].j / (double)javabin[i].f;            }        }        qsort(0,n-1,javabin);        double total = 0;        for (int i = n-1;i >= 0;i--){            if (javabin[i].bili == (double)(javabin[i].j * 10000)){                javabin[i].bili = javabin[i].bili / 10000;                total = total + javabin[i].bili;            }            else if (m >= javabin[i].f){                total = total + javabin[i].f * javabin[i].bili;                m = m - javabin[i].f;                if (m == 0){//刚好分配完;                    break;                }            }            else{                total = total + m * javabin[i].bili;                break;            }        }        printf("%.3lf\n", total);    }}int main(){    test();    return 0;}

注意：此题需要优先考虑不用猫粮就可以兑换JavaBean的情况（此时除数为零）（我的方法是将j直接作为比例，然后扩大）。