递归题：(在有inc()和dec()的情况下)不允许用==以外的运算符和for()/while()计算两个数的乘积
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今天这个题目有点意思。
题目要求：计算两个数的乘积
限制1：不允许使用==以外的运算符（除了定义inc()和dec()外）
限制2：不允许使用for和while
限制3：不允许导入各种奇怪的函数
这题就用递归做啦~
这道题目恶心之处在于不允许用==以外的运算符导致我想了5分钟很久。嗯。
首先还是定义inc和dec。
def inc(n): return n+1; def dec(n): return n-1;
然后定义add(a,b)。效果是计算a+b的值。这个就用递归把b拆成0+1+1+...+1(一共b个1)就行了。
def add(a,b): if (b == 0): return a; else: return inc(add(a,dec(b)));
之后就定义一个mul(a,b)。效果计算a*b的值。和上面一个原理一样，把a*b拆成0+a+a+...+a(一共b个a)就行了。
代码见下。
def mul(a,b): if (b == 0): return 0; else: return add(a,mul(a,dec(b)));
最后在main里把I/O处理掉就结束。
print(mul(int(input()),int(input())));
（未完待续？）
吐槽一句——这是哪个老师出的这种神奇题目 就算练递归也不需要这么练吧= =
2017年11月29日 14:36
杰卡森诈尸啦
2017年12月04日 15:42
日常巡视JCW的blog
你可以尝试做一下 a+b problem 全程禁止使用加号和减号
甚至我们可以考虑更进一步,我们禁止你用乘号和除号,再加个if也可以
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