问题描述如下:
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
算法分析:
该算法采用先对兑换比例进行排序;然后再从比率最高的进行兑换;直到兑换完为止。
参考代码:
public static void main(String[] args) {
// TODO Auto-generated method stub
int m, n, i, j, tempa, tempb;
float temp;
float sum = 0;
Scanner scanner = new Scanner(System.in);
System.out.println("请输入有多少猫粮:");
m = scanner.nextInt();
System.out.println("请输入有多少房间:");
n = scanner.nextInt();
System.out.println("猫粮:" + m + "\t房间:" + n);
int a[] = new int[n];
int b[] = new int[n];
float r[] = new float[n];
System.out.println("依次输入可兑换的javabean,和需要的猫粮:");
for (i = 0; i < n; i++) {
a[i] = scanner.nextInt();
b[i] = scanner.nextInt();
r[i] = (float) a[i] / b[i];
}
for (i = 0; i < n; i++) {
System.out.print(a[i] + "---");
System.out.println(b[i]);
}
for (i = 0; i < n - 1; i++) {
for (j = 0; j < n - i - 1; j++)
if (r[j] < r[j + 1]) {
temp = r[j];
r[j] = r[j + 1];
r[j + 1] = temp;
tempa = a[j];
a[j] = a[j + 1];
a[j + 1] = tempa;
tempb = b[j];
b[j] = b[j + 1];
b[j + 1] = tempb;
}
}
for (i = 0; i < n; i++) {
System.out.print(r[i] + "---");
System.out.print(a[i] + "---");
System.out.println(b[i]);
}
if (m > 0) {
for (i = 0; i < b.length; i++) {
if (m >= b[i]) {
sum += a[i];
m = m - b[i];
} else {
sum += (float) m / b[i] * a[i];
break;
}
}
} else {
System.out.println("兑换结束!!!!");
}
System.out.println(sum);
}