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  • BZOJ 4816: [Sdoi2017]数字表格

    二次联通门 : BZOJ 4816: [Sdoi2017]数字表格

    /*
        BZOJ 4816: [Sdoi2017]数字表格
    
        莫比乌斯反演
    
        妈呀,我的μ没有啦
        
    */
    #include <cstdio>
    #include <iostream>
    #define rg register
    #define Max 1000050
    #define mo 1000000007
    inline int min (int a, int b) { return a < b ? a : b; }
    inline void read (int &n)
    {
        rg char c = getchar ();
        for (n = 0; !isdigit (c); c = getchar ());
        for (; isdigit (c); n = n * 10 + c - '0', c = getchar ());
    }
    typedef long long LL;
    LL f[Max], invf[Max];
    
    LL pow (LL x, LL y)
    {
        LL res = 1;
        for (; y; y >>= 1, (x *= x) %= mo)
            if (y & 1) (res *= x) %= mo;
        return res;
    }
    inline void pls (LL &a, LL b) { a += b; if (a >= mo) a -= mo; }
    int main (int argc, char *argv[])
    {
        rg int i, j; LL inv; int N, M, T;
        f[1] = 1;
        for (i = 2; i < Max; ++ i) pls (f[i], f[i - 1] + f[i - 2]);
        for (i = 1; i < Max; ++ i)
        {
            inv = pow (f[i], mo - 2);
            for (j = i << 1; j < Max; j += i) (f[j] *= inv) %= mo;
        }
        for (f[0] = i = 1; i < Max; ++ i) (f[i] *= f[i - 1]) %= mo;
        for (i = 0; i < Max; ++ i) invf[i] = pow (f[i], mo - 2);
    
        read (T); LL Answer;
        for (; T; -- T)
        {
            read (N), read (M); Answer = 1;
            if (N > M) std :: swap (N, M);
            for (i = 1; i <= N; i = j + 1)
            {
                j = min (N / (N / i), M / (M / i));
                (Answer *= pow (pow (f[j] * invf[i - 1] % mo, (LL) N / i), (LL) M / i)) %= mo;
            }
            printf ("%lld
    ", Answer);
        }    
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/ZlycerQan/p/8074665.html
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