FatMouse' Trade 解题报告

原题： 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 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