鸡兔同笼-数学解法

今有雉兔同笼,上有三十五头,下有九十四足,问雉兔各几何?
一般解法:
穷举
public void chr() {
    for (int i = 0; i <= 35; i++) {
        if (i * 2 + (35 - i) * 4 == 94) {
            System.out.println("笼子里有雉[" + i + "]兔[" + (35 - i) + "]");
        }
    }
}

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数学解法:
一元二次方程:
设 鸡为x,兔为y
{
x + y = 35
2x + 4y = 94
}
 
x = 35 - y
2x = 94 - 4y
::x = 47 - 2y = 35 - y
y = 47 -35 = 12;
 
x = 35 - y = 35 - 12 = 23
 
x = 23,y = 12
 
提炼规律:
47 - 2y = 35 - y
y = 47 - 35
y = 脚 / 2 - 头
public void chr1() {
    int head = 35;
    int foot = 94;
    int x, y;
    y = foot / 2 - head;
    x = 35 - y;
    System.out.println("笼子里有雉[" + x + "]兔[" + y + "]");
}
 
原文地址:https://www.cnblogs.com/chenglc/p/10963091.html