Hard nim

https://vjudge.net/problem/黑暗爆炸-4589
FWT模板题

struct FWT{
    const int inv2  = ksm(2, mod - 2);
    int l;
    void init(int n) {
        l = 2;
        while(l < n)    l <<= 1;//l <= n?
    }
    inline int qmod(int x) {
        while(x >= mod) x -= mod;
        return x;
    }
    //第一行for(int k = n; k > 1; k >>= 1) 与 for(int k = 2; k <= n; k <<= 1)   等价，无影响 -> k的枚举顺序可以任意调换。
    //从大到小枚举复杂度更优？
    void fwt(vector<int> &a, int opt, int cas) {
        a.resize(l);//l + 1?
        int x, y;
        for(int k = 2; k <= l; k <<= 1) {
            int mid = k >> 1;
            for(int i = 0; i < l; i += k) {
                for(int j = i, up = i + mid; j < up; ++ j) {
                    //or:
                    if(cas == 1)    a[j + mid] = qmod(a[j + mid] + opt * a[j] + mod);

                    //and:
                    else if(cas == 2)   a[j] = qmod(a[j] + opt * a[j + mid] + mod);

                    //xor:
                    else if(cas == 3)   {
                        x = a[j], y = a[j + mid];
                        a[j] = qmod(x + y);
                        a[j + mid] = qmod(x - y + mod);
                        if(opt == -1) a[j] = 1ll * a[j] * inv2 % mod, a[j + mid] = 1ll * a[j + mid] * inv2 % mod;
                    }

                    //xnor:
                    else {
                        x = a[j], y = a[j + mid];
                        a[j] = qmod(y - x + mod);
                        a[j + mid] = qmod(x + y);
                        if(opt == -1)  a[j] = 1ll * a[j] * inv2 % mod, a[j + mid] = 1ll * a[j + mid] * inv2 % mod;
                    }
                }
            }
        }
    }
}f;

//FMT的常数比FWT小一点
struct FMT{
    const int inv2 = ksm(2, mod - 2);
    int l;
    void init(int n) {
        l = 2;
        while(l < n)    l <<= 1;
    }
    inline int qmod(int x) {
        while(x >= mod) x -= mod;
        return x;
    }
    void fmt(vector<int> &a, int opt, int cas) {
        a.resize(l);
        int x, y;
        for(int k = 1; k < l; k <<= 1) {
            //or:
            if(cas == 1)    {
                for(int i = 0; i < l; ++ i) if(~i & k)  a[i | k] = qmod(a[i | k] + opt * a[i] + mod);
            }

            //and:
            else if(cas == 2)   {
                for(int i = l - 1; i >= 0; -- i) if(i & k)  a[i ^ k] = qmod(a[i ^ k] + opt * a[i] + mod);
            }

            //xor:
            else {
                for(int i = 0; i < l; ++ i) {
                    if(~i & k)  {
                        x = a[i], y = a[i | k];
                        a[i] = qmod(x + y);
                        a[i | k] = qmod(x - y + mod);
                        if(opt == -1)   a[i] = 1ll * a[i] * inv2 % mod, a[i | k] = 1ll * a[i | k] * inv2 % mod;
                    }
                }
            }

        }
    }
}f2;

int a[maxn];
void run() {
    int n, m;
    for(int i = 2; i < maxn; ++ i)    a[i] = 1;
    for(int i = 2; i < maxn; ++ i) if(a[i])
        for(int j = i + i; j < maxn; j += i)  a[j] = 0;
    while(~scanf("%d %d", &n, &m)) {
        //n堆石子，每堆个数不超过m的素数
        //f_{i ^ j = 0} = a_i * a_j//a_i = 1(i是素数)
        vector<int> b(m + 1);
        for(int i = 0; i <= m; ++ i)    b[i] = a[i];
        f2.init(m + 1);
        f2.fmt(b, 1, 3);
        for(int i = 0; i < f2.l; ++ i) b[i] = ksm(b[i], n);//每个数可以选n次
        f2.fmt(b, -1, 3);
        printf("%d\n", b[0]);
    }
    return ;
}