1 条题解

  • 1
    @ 2024-9-14 16:12:33

    参考答案:

    #include<iostream>
    using namespace std;
    typedef long long LL;
    const int N = 1e6 + 10;
    LL phi[N], primes[N], st[N];
    LL n;
    LL cnt, res;
    void f(LL n)
    {
        phi[1] = 1;
        for (int i = 2; i <= n; ++i)
        {
            if (!st[i])
            {
                primes[++cnt] = i;
                phi[i] = i - 1;
            }
            for (int j = 1; primes[j] <= n / i; ++j)
            {
                st[primes[j] * i] = 1;
                if (i % primes[j] == 0)
                {
                    phi[primes[j] * i] = primes[j] * phi[i];
                    break;
                }
                else phi[primes[j] * i] = phi[i] * (primes[j] - 1);
            }
        }
    }
    int main()
    {
        cin >> n;
        f(n);
        for (int i = 1; i <= n; ++i) res += phi[i];
        cout << res << endl;
        return 0;
    }
    
    • 1

    信息

    ID
    5367
    时间
    1000ms
    内存
    256MiB
    难度
    10
    标签
    递交数
    3
    已通过
    2
    上传者