《Cracking the Coding Interview》——第7章：数学和概率论——题目4

2014-03-20 02:16

题目：只用加法和赋值，实现减法、乘法、除法。

解法：我只实现了整数范围内的。减法就是加上相反数。乘法就是连着加上很多个。除法就是减到不能减为止，数数总共减了多少个。

代码：

// 7.4 Implement the -*/ function with only the + operator.
// You cannot use bit operation, although you might want it for efficiency.
#include <cstdio>
using namespace std;

int negate(int n)
{
    int one = (n >= 0 ? -1 : 1);
    int res = 0;
    
    while (n != 0) {
        n += one;
        res += one;
    }
    
    return res;
}

int add(int x, int y)
{
    return x + y;
}

int subtract(int x, int y)
{
    return x + negate(y);
}

int multiply(int x, int y)
{
    int res = 0;
    int value = (x >= 0 ? y : negate(y));
    int one = (x >= 0 ? -1 : 1);
    
    while (x != 0) {
        res += value;
        x += one;
    }
    
    return res;
}

int divide(int x, int y)
{
    if (y == 0) {
        return 0;
    }
    if (x == 0) {
        return 0;
    }
    
    int res = 0;
    int one = 1;
    int negone = -1;
    int negy = negate(y);
    
    if (x > 0) {
        if (y > 0) {
            while (x >= y) {
                x += negy;
                res += one;
            }
        } else {
            while (x >= negy) {
                x += y;
                res += negone;
            }
        }
    } else {
        if (y > 0) {
            while (x <= negy) {
                x += y;
                res += negone;
            }
        } else {
            while (x <= y) {
                x += negy;
                res += one;
            }
        }
    }
    
    return res;
}

int main()
{
    char s[10];
    int a, b;
    int res;
    
    while (scanf("%d%s%d", &a, s, &b) == 3) {
        switch (s[0]) {
        case '+':
            res = add(a, b);
            break;
        case '-':
            res = subtract(a, b);
            break;
        case '*':
            res = multiply(a, b);
            break;
        case '/':
            res = divide(a, b);
            break;
        }
        printf("%d\n", res);
    }
    
    return 0;
}