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 = , MO = 1e9 + , inv2 = (MO + ) / ;

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

 inline void FWT_or(int *a, int n, int f) {
     for(int len = ; len < n; len <<= ) {
         for(int i = ; i < n; i += (len << )) {
             for(int j = ; 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 = ; len < n; len <<= ) {
         for(int i = ; i < n; i += (len << )) {
             for(int j = ; 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 = ; len < n; len <<= ) {
         for(int i = ; i < n; i += (len << )) {
             for(int j = ; 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 == -) {
                     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 = , x; i <= n; i++) {
         scanf("%d", &x);
         a[x]++;
     }
     n = ;
     lm =  << ;
     cnt[] = f[] = ;
     for(int i = ; i < lm; i++) {
         f[i] = (f[i - ] + f[i - ]) % MO;
         cnt[i] = cnt[i - (i & (-i))] + ;
     }
     memcpy(c, a, lm * sizeof(int));

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

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

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

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

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

     printf("%d\n", ans);
     return ;
 }

AC代码