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  • 求 1! + 2! + 3! + ... + n!

    long long unsigned factorial(long long unsigned num)
    {
        return num < 2LL? 1LL : (factorial(num - 1LL) * num);
    
    }
    
    long long unsigned factoricalSum(long long unsigned num)
    {
        return num < 2LL? 1LL : factoricalSum(num - 1LL) + factorial(num);
    }
    long long unsigned factoricalSum(long long unsigned num)
    {
        function<long long unsigned(long long unsigned)> factorial;
        factorial = [&factorial](decltype(num) n){return n < 2LL? 1LL : (factorial(n - 1LL) * n);};
        return num < 2LL? 1LL : factoricalSum(num - 1LL) + factorial(num);
    }

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  • 原文地址:https://www.cnblogs.com/wuOverflow/p/4653013.html
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