FatMouse' Trade 解题报告
2013-01-22 13:33
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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
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