Question

问题1的简单解决方案是

static unsigned int solutionInefficient(unsigned int n){
    unsigned int sum = 0;
    for (unsigned int i = 0; i < n; i++){
        if (i % 3 == 0 || i % 5 == 0) {
            sum += i;
        }
    }
    return sum;
}

我决定尝试使用n = 2147483647的不同测试用例，最终结果在12秒内计算出来。所以，我提出了另一种解决方案，它给了我相同的答案并花了2秒钟：

static unsigned int solutionEfficient(unsigned int n){
    unsigned int sum = 0;
    unsigned int sum3 = 0;
    unsigned int sum5 = 0;
    unsigned int sum15 = 0;
    for (unsigned int i = 3; i < n; i += 3){
        sum3 += i;
    }

    for (unsigned int i = 5; i < n; i += 5){
        sum5 += i;
    }
    for (unsigned int i = 15; i < n; i += 15){
        sum15 += i;
    }
    return sum3 + sum5 - sum15;
}

我最后一次尝试更快地实现了一些谷歌搜索和使用算术求和公式，最后一段代码看起来像这样：

static unsigned int solutionSuperEfficient(unsigned int n){
    n = n - 1;
    unsigned int t3 = n / (unsigned int)3,
        t5 = n / (unsigned int)5,
        t15 = n / (unsigned int)15;
    unsigned int res_3 = 3 * (t3 * (t3 + 1)) *0.5;
    unsigned int res_5 = 5 * (t5 * (t5 + 1)) *0.5;
    unsigned int res_15 = 15 * (t15 * (t15 + 1)) *0.5;

    return res_3 + res_5 - res_15;
}

然而，这并未为此测试案例提供正确的答案。它确实为n = 1000提供了正确答案。我不确定为什么我的测试用例失败了，有什么想法？

Answer 1

您的超高效解决方案存在两个问题：

您使用的是浮点数0.5而不是除以2.这将导致舍入错误。请注意，保证x *（x + 1）是偶数，因此您可以安全地除以2。
整数溢出。计算t3 * (t3 + 1)和类似产品将溢出unsigned int。为避免这种情况，请改用unsigned long long。

以下是更正后的代码：

static unsigned int solutionSuperEfficient(unsigned int n){
    n = n - 1;
    unsigned long long t3 = n / 3,
        t5 = n / 5,
        t15 = n / 15;
    unsigned long long res_3 = 3ULL * ((t3 * (t3 + 1)) / 2ULL);
    unsigned long long res_5 = 5ULL * ((t5 * (t5 + 1))  / 2ULL);
    unsigned long long res_15 = 15LL * ((t15 * (t15 + 1)) / 2ULL);

    return (unsigned int)(res_3 + res_5 - res_15);
}

实际上，您不需要t3，t5和t15为unsigned long long，因为这些值永远不会溢出unsigned int。

项目Euler＃1测试扩展

1 个答案: