Question

int add2_recurse(int a, int b) { // recursive

// Rule: Can't use the *, /, +, =, *=, /=, +=, -= operators.

// Can use: ++ and/or --

// No loops allowed; no static local or global variables.



// What I have so far

while(b--) {
    a++;
return a; }

int main() {
show_test(4, "add2", recursive);
cout << add2_recurse(4,5); cout<<" ";  // correct:   9
cout<<add2_recurse(-5, 15); cout<<" ";  // correct:  10
cout<<add2_recurse(20, -9); cout<<" ";  // correct:  11
cout<<add2_recurse(-7, -5); cout<<" ";  // correct: -12
cout<<endl;

当我运行时，＆＃34; 9 10 11 -12＆＃34;的正确输出显示时，只有11和-12显示得慢得多。关于如何让它运行得更快的任何想法？

Answer 1

首先，您的解决方案有两个原因是错误的。您不使用递归，并且您正在使用循环。

至于为什么它在最后两种情况下运行得如此之慢：只要b为负，b一直递减到最小可能的整数，然后它会回绕到最大值可能的整数，最后它会递减直到它为0.假设一个32位整数，你将有大约40亿次迭代的循环。

您需要区分b的负值和正值，然后根据需要递减或递增a。

Answer 2

你不需要任何递归，在按位运算方面很容易实现加法：

#include <limits.h> /* CHAR_BIT for pedantic correctness */

int add(int a_, int b_)
{
    /*
    Signed types tend to have UB. Prevent this by using unsigned.
    This might only work with twos-complement systems, sorry - patches welcome.
    */
    const unsigned a = (unsigned)a_;
    const unsigned b = (unsigned)b_;
    unsigned s = 0, c = 0;
    int i;

    /* constant-sized loop, can be trivially unrolled if you know how many bits are in an int (e.g. 16 or 32) */
    for (i = 0; i < sizeof(int) * CHAR_BIT; ++i)
    {
        s |= ((a ^ b ^ c) & (1 << i));
        c = ((a & b) | (a & c) | (b & c)) & (1 << i);
        c <<= 1;
    }

    return (signed)s;
}

谨防上述代码中的错误;我只是证明它是正确的，没有尝试过。

如何使这个递归函数运行得更快？

2 个答案: