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  • 数学--数论--HDU 2582 F(N) 暴力打表找规律

    This time I need you to calculate the f(n) . (3<=n<=1000000)

    f(n)= Gcd(3)+Gcd(4)+…+Gcd(i)+…+Gcd(n).
    Gcd(n)=gcd(C[n][1],C[n][2],……,C[n][n-1])
    C[n][k] means the number of way to choose k things from n some things.
    gcd(a,b) means the greatest common divisor of a and b.
    Input
    There are several test case. For each test case:One integer n(3<=n<=1000000). The end of the in put file is EOF.
    Output
    For each test case:
    The output consists of one line with one integer f(n).
    Sample Input
    3
    26983
    Sample Output
    3
    37556486
    题目就是这么短小精悍,这题我实在不知道怎么写,然后也不会数论的推理,我就打了表,发现跟质数的关系很大,就顺着推了一下, 就过了。

    #include <iostream>
    #include<cstdio>
    #define ll long long
    #define maxn 1000000
    using namespace std;
    int zs[maxn], t = 0, n;
    ll f[maxn + 5];
    bool v[maxn + 5];
    
    int main()
    {
        for (int i = 2; i <= maxn; i++)
        {
            if (!v[i]) f[i] = zs[++t] = i;
            for (int j = 1,u = zs[j] * i;  j <= t && u<= maxn; j++)
            {
                v[u] = 1;
                if (!(i % zs[j]))
                {
                    f[u] = f[i];
                    break;
                }
                f[u] = 1;
                u = zs[j+1] * i; 
            }
        }
        for (int i = 4; i <= maxn; i++)
            f[i] += f[i - 1];
        while (scanf("%d", &n)!=EOF)
            printf("%lld
    ", f[n]);
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/lunatic-talent/p/12798498.html
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