题目描述:
"嘟嘟嘟嘟嘟嘟
喇叭响
我是汽车小司机
我是小司机
我为祖国运输忙
运输忙"
这是儿歌“快乐的小司机”。话说现在当司机光有红心不行，还要多拉快跑。多拉不是超载，是要让所载货物价值最大，特别是在当前油价日新月异的时候。司机所拉货物为散货，如大米、面粉、沙石、泥土......
现在知道了汽车核载重量为w，可供选择的物品的数量n。每个物品的重量为gi,价值为pi。求汽车可装载的最大价值。（n<10000,w<10000,0<gi<=100,0<=pi<=100)
代码:
package lanqiao;import java.util.*;public class Main {public static void main(String[] args) {Scanner sc = new Scanner(System.in);int n = sc.nextInt(); //数量int w = sc.nextInt(); //载重double[] gi = new double[n]; //重量double[] pi = new double[n]; //价值double[] arr = new double[n];for(int i = 0;i < pi.length;i ++){gi[i] = sc.nextDouble();pi[i] = sc.nextDouble();arr[i] = pi[i] / gi[i];}for(int i = 0;i < arr.length - 1;i ++){for(int j = 0;j < arr.length - i - 1;j ++){//排序if(arr[j]<arr[j+1]){double g =gi[j];gi[j]=gi[j+1];gi[j+1]=g;double p =pi[j];pi[j]=pi[j+1];pi[j+1]=p;double a =arr[j];arr[j]=arr[j+1];arr[j+1]=a;}}}double sum = 0;for(int i = 0;i < arr.length;i ++){if(w >= gi[i]){sum += pi[i];w -= gi[i];}else{sum += w * arr[i];break;}}System.out.printf("%.1f",sum);}
}
)
)
