Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
题目链接
http://acm.hdu.edu.cn/showproblem.php?pid=1009
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 31.50 总结
初步了解了贪心算法，是选择价值比最大的东西来拿（由于可以切开拿，不是必须整体），接下来就好多了。
遇到的问题主要是浮点数的误差，只能用价值比（浮点数）比较大小，后面的运算求和时不能用它相乘来计算，拿样例来说，结果是13.333，而我没改进前，算出来是12.444.可见是误差。
第二个就是网上其他人使用结构体来储存，这样排序时其他变量能一起移动，但我开了3个数组，而且也不能用sort来排序了，只能自己写了选排。
#include<iostream> #include<cstdio> using namespace std; double price[1005]; int num1[1005],num2[1005]; void sort1(double* a,int x) { int b; double c; for(int i=0;i<x;i++){ for(int j=i+1;j<x;j++){ if(a[i]<a[j]){ c=a[i]; a[i]=a[j]; a[j]=c; b=num1[i]; num1[i]=num1[j]; num1[j]=b; b=num2[i]; num2[i]=num2[j]; num2[j]=b; } } } } int main() { int m,n,j; while(scanf("%d %d",&m,&n)==2){ if(m==-1&&n==-1) break; for(int i=0;i<n;i++){ scanf("%d %d",&num2[i],&num1[i]); price[i]=(double)num2[i]/num1[i]; } sort1(price,n); //for(int i=0;i<n;i++) cout <<price[i]<<num[i]<<endl; double sum=0; for(int i=0;i<n;i++){ if(m<=0) break; if(m>=num1[i]){ sum+=num2[i]; m-=num1[i]; } else { sum+=(double)m/num1[i]*num2[i]; m=0; } } printf("%.3lf\n",sum); } return 0; }
转载于:https://www.cnblogs.com/wangdongkai/p/5267451.html
相关资源：数据结构—成绩单生成器