FatMouse' Trade 解题报告

原题：Problem DescriptionFatMouse 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. InputThe 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. OutputFor 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 Input5 37 24 35 220 325 1824 1515 10-1 -1 Sample Output13.33331.500分析：贪心算法——找最优解，我的思路是，运用结构体，运用sort函数将数据从大到小排序，先换: 7  2; 再换 5 2；最后用剩下的1个，换1.33的javabeen。，然后问题就迎刃而解了。。源码：#include<algorithm>#include<iostream>#include<stdio.h>using namespace std;//一些c++头文件，刚刚跟师哥学习的struct node{    int x;    int y;    double z;}p[1005];//this is结构体int cmp(node a,node b){    return a.z>b.z;}//不加这部分默认是从小到大排列int main(){    int m,n,i;    double s;    while(scanf("%d%d",&m,&n)&&(m!=-1||n!=-1))    {        s=0.0;        for(i=0; i<n; i++)        {            scanf("%d%d",&p[i].x,&p[i].y);            p[i].z=p[i].x*1.0/p[i].y;        }        sort(p,p+n,cmp);//记住这个格式————排序；        for(i=0; i<n; i++)        {            if(m>=p[i].y)            {                s=s+p[i].x;                m=m-p[i].y;            }            else            {                s=s+m*p[i].z;                break;            }        }        printf("%.3f\n",s);    }    return 0;}//OK