CF914G Sum the Fibonacci

解：发现我们对a和b做一个集合卷积，对d和e做一个^FWT，然后把这三个全部对位乘上斐波那契数，然后做&FWT就行了。

#include <bits/stdc++.h>

const int N = 150010, MO = 1e9 + 7, inv2 = (MO + 1) / 2;

int n, lm, f[N], a[N], b[N], c[N], cnt[N], d[20][N], e[20][N];

inline void FWT_or(int *a, int n, int f) {
    for(int len = 1; len < n; len <<= 1) {
        for(int i = 0; i < n; i += (len << 1)) {
            for(int j = 0; j < len; j++) {
                a[i + len + j] = ((a[i + len + j] + f * a[i + j]) % MO + MO) % MO;
            }
        }
    }
    return;
}

inline void FWT_and(int *a, int n, int f) {
    for(int len = 1; len < n; len <<= 1) {
        for(int i = 0; i < n; i += (len << 1)) {
            for(int j = 0; j < len; j++) {
                a[i + j] = ((a[i + j] + f * a[i + len + j]) % MO + MO) % MO;
            }
        }
    }
    return;
}

inline void FWT_xor(int *a, int n, int f) {
    for(int len = 1; len < n; len <<= 1) {
        for(int i = 0; i < n; i += (len << 1)) {
            for(int j = 0; j < len; j++) {
                int t = a[i + len + j];
                a[i + len + j] = (a[i + j] - t + MO) % MO;
                a[i + j] = (a[i + j] + t) % MO;
                if(f == -1) {
                    a[i + len + j] = 1ll * a[i + len + j] * inv2 % MO;
                    a[i + j] = 1ll * a[i + j] * inv2 % MO;
                }
            }
        }
    }
    return;
}

int main() {
    int n;
    scanf("%d", &n);
    for(int i = 1, x; i <= n; i++) {
        scanf("%d", &x);
        a[x]++;
    }
    n = 17;
    lm = 1 << 17;
    cnt[1] = f[1] = 1;
    for(int i = 2; i < lm; i++) {
        f[i] = (f[i - 1] + f[i - 2]) % MO;
        cnt[i] = cnt[i - (i & (-i))] + 1;
    }
    memcpy(c, a, lm * sizeof(int));

    for(int i = 0; i < lm; i++) {
        d[cnt[i]][i] = a[i];
    }
    for(int i = 0; i <= n; i++) {
        FWT_or(d[i], lm, 1);
    }
    for(int i = 1; i <= n; i++) {
        for(int j = 0; j <= i; j++) {
            for(int s = 0; s < lm; s++) {
                e[i][s] = (e[i][s] + 1ll * d[j][s] * d[i - j][s] % MO) % MO;
            }
        }
    }
    for(int i = 0; i <= n; i++) {
        FWT_or(e[i], lm, -1);
    }
    for(int i = 0; i < lm; i++) {
        b[i] = 1ll * e[cnt[i]][i] * f[i] % MO;
    }

    FWT_xor(c, lm, 1);
    for(int i = 0; i < lm; i++) {
        c[i] = 1ll * c[i] * c[i] % MO;
    }
    FWT_xor(c, lm, -1);
    for(int i = 0; i < lm; i++) {
        c[i] = 1ll * c[i] * f[i] % MO;
    }

    for(int i = 0; i < lm; i++) {
        a[i] = 1ll * a[i] * f[i] % MO;
    }

    FWT_and(a, lm, 1);
    FWT_and(b, lm, 1);
    FWT_and(c, lm, 1);
    for(int i = 0; i < lm; i++) {
        a[i] = 1ll * a[i] * b[i] % MO * c[i] % MO;
    }
    FWT_and(a, lm, -1);

    int ans = 0;
    for(int i = 1; i < lm; i <<= 1) {
        ans = (ans + a[i]) % MO;
    }

    printf("%d
", ans);
    return 0;
}