莫比乌斯反演初步与实际应用

骑猪看日落 2022-03-07 13:26 278阅读 0赞

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#include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    using namespace std ;
    
    typedef long long LL ;
    const int MAXN = 10000010 ;
    
    long long T, N, M, V[MAXN], P[MAXN], Mu[MAXN], Tot ;
    long long S[MAXN], Ans, G[MAXN] ;
    
    inline long long  Read() {
        long long X = 0, F = 1 ; char ch = getchar() ;
        while (ch > '9' || ch < '0') F = (ch == '-' ? - 1 : 1), ch = getchar() ;
        while (ch >= '0' && ch <= '9') X=(X<<1)+(X<<3)+(ch^48), ch = getchar() ;
        return X * F ;
    }
    
    inline void MU() {
        memset(V, 0, sizeof(V)) ;
        Mu[1] = 1 ; Tot = 0 ;
        for (int i = 2 ; i < MAXN ; i ++) {
            if (! V[i]) P[Tot ++] = i, Mu[i] = - 1, G[i] = 1 ;
            for (int j = 0 ; j < Tot && i * P[j] < MAXN ; j ++) {
                V[i * P[j]] = 1 ;
                if (i % P[j]) Mu[i * P[j]] = - Mu[i],
                    G[i * P[j]] = Mu[i] - G[i] ; else {
                    Mu[i * P[j]] = 0 ; G[i * P[j]] = Mu[i] ;
                    break ;
                }
            }
        }
        for (int i = 1 ; i < MAXN ; i ++) S[i] = S[i - 1] + G[i] ;
    }
    
    int main() {
        MU() ;
        T = Read() ; while (T --) { 
            Ans = 0 ;
            N = Read(), M = Read() ; 
            for (int i = 1, j ; i <= min(N, M) ; i = j + 1) {
                j = min(N / (N / i), M / (M / i)) ;
                Ans += (N / i) * (M / i) * (S[j] - S[i - 1]) ;
            }
            printf("%lld\n", Ans) ;
        }
    }

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[clipboard.png 1]: /images/20220307/15a4a4b8225e4127b8822f48f7af4527.png
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