GO语言百钱百鸡

package main

//公鸡5钱1只，母鸡3钱一只，小鸡1钱三只 需要用百钱买百鸡
//100钱的话 公鸡20或者母鸡33 小鸡100
func main() {
    count := 0
    for i := 0; i <= 20; i++{
        for j := 0; j <= 33; j++ {
            for k := 0; k <= 100; k++ {
                if i+j+k == 100 && i*5+j*3+k/3 == 100 && k%3 == 0 {
                    println(i, j, k)
                }
                count++
            }
        }
    }
    println(count)

    count = 0
    //优化1
    for i := 0; i <= 20; i++{
        for j := 0; j <= 33; j++ {
            for k := 0; k <= 100; k+=3 {
                if i+j+k == 100 && i*5+j*3+k/3 == 100 {
                    println(i, j, k)
                }
                count++
            }
        }
    }
    println(count)
    
    count = 0
    println()
    //优化2 总共一百只鸡 100-公鸡-母鸡等于小鸡 只需要判断百钱
    for i := 0; i <= 20; i++{
        for j := 0; j <= 33; j++ {
            k := 100 - i - j
            if k % 3 == 0 && i*5+j*3+k/3 == 100{
                println(i, j, k)
            }
            count++
        }
    }
    println(count)

    //优化3 k每次加三个 就避免了判断k%3==0, j<0 表明公鸡小鸡买的太多了，直接进入下一次迭代减少执行次数
    count = 0
    for i := 0; i <= 20; i++{
        for k := 0; k <= 100; k += 3 {
            j := 100 - i - k
            if j < 0 {
                continue
            }
            if i*5+j*3+k/3 == 100{
                println(i, j, k)
            }
            count++
        }
    }
    println(count)
}

优化三跑得快 输出如下

0 25 75
4 18 78
8 11 81
12 4 84
72114
0 25 75
4 18 78
8 11 81
12 4 84
24276

0 25 75
4 18 78
8 11 81
12 4 84
714
0 25 75
4 18 78
8 11 81
12 4 84
644