一道偏结论的题。当 (n=2^{k+1}) 时
[sum_{i=0, ext{popcount}(i)equiv 0pmod 2}^{2^{k+1}-1}i^k=sum_{i=0, ext{popcount}(i)equiv 1pmod 2}^{2^{k+1}-1}i^k
]
采用归纳法证明（后文为方便以 (=0/1) 表示 (equiv 0/1)）。当 (k=0) 时，结论显然成立；
当 (k>0) 时，欲证明结论，先移项：
[sum_{i=0, ext{popcount}(i)=0}^{2^{k+1}-1}i^k-sum_{i=0, ext{popcount}(i)=1}^{2^{k+1}-1}i^k=0
]
对于前缀相同的二进制 ((cdots 0)) 和 ((cdots 1)) 均出现，仅最低位不同，考虑这样枚举
[sum_{i=0}^{2^k-1}c_i imes left((2i+1)^k-(2i)^k
ight)=0
]
(c_i) 为 (+1) 或 (-1)，取决于前缀 (i) 中 1 的个数。不难发现 (c_i=(-1)^{ ext{popcount}(i)})
所以得到：
[sum_{i=0}^{2^k-1}(-1)^{ ext{popcount}(i)}left((2i+1)^k-(2i)^k
ight)=0
]
展开得到：
[sum_{i=0}^{2^k-1}(-1)^{ ext{popcount}(i)}sum_{j=0}^{k-1}{kchoose j}(2i)^j=0
]
对于 ( ext{popcount}(i)) 分组，就变成：
[sum_{j=0}^{k-1}2^j{kchoose j}sum_{i=0, ext{popcount}(i)=0}^{2^k-1}i^j=sum_{j=0}^{k-1}2^j{kchoose j}sum_{i=0, ext{popcount}(i)=1}^{2^k-1}i^j (*)
]
由归纳，当 (k'<k) 时：
[sum_{i=0, ext{popcount(i)}=0}^{2^{k'+1}-1}i^{k'}=sum_{i=0, ext{popcount(i)}=1}^{2^{k'+1}-1}i^{k'}
]
引理：
[sum_{i=0, ext{popcount(i)}=0}^{t imes 2^{k'+1}-1}i^{k'}=sum_{i=0, ext{popcount(i)}=1}^{t imes 2^{k'+1}-1}i^{k'}, tin
]
证明引理，我们只需要证明下式：
[sum_{i=t imes 2^{k'+1}, ext{popcount}(i)=0}^{(t+1) imes 2^{k'+1}-1}i^{k'}=sum_{i=t imes 2^{k'+1}, ext{popcount}(i)=1}^{(t+1) imes 2^{k'+1}-1}i^{k'}
]
我们同样可以归纳，这里省略掉部分细节。改写一下：
[sum_{i=0, ext{popcount}(i)=0}^{2^{k'+1}-1}(i+t imes 2^{k'+1})^{k'}=sum_{i=0, ext{popcount}(i)=1}^{2^{k'+1}-1}(i+t imes 2^{k'+1})^{k'}
]
同样展开：
[sum_{i=0, ext{popcount}(i)=0}^{2^{k'+1}-1}sum_{j=0}^{k'}{k'choose j}i^j(t imes 2^{k'+1})^{k'-j}=sum_{i=0, ext{popcount}(i)=1}^{2^{k'+1}-1}sum_{j=0}^{k'}{k'choose j}i^j(t imes 2^{k'+1})^{k'-j}
]
[Rightarrowsum_{j=0}^{k'}(t imes 2^{k'+1})^{k'-j}{k'choose j}sum_{i=0, ext{popcount}(i)=0}^{2^{k'+1}-1}i^j=sum_{j=0}^{k'}(t imes 2^{k'+1})^{k'-j}{k'choose j}sum_{i=0, ext{popcount}(i)=1}^{2^{k'+1}-1}i^j
]
由归纳假设每一项都相等，由此引理得证。
由上，对比每一项，能直接得到 ((*)) 式成立。故结论成立
推导过程中，我们甚至能知道
[sum_{i=0, ext{popcount}(i)=0}^{t imes 2^{k+1}-1}i^k=sum_{i=0, ext{popcount}(i)=1}^{t imes 2^{k+1}-1}i^k
]
成立。这样我们就可以将复杂度从 (mathcal O(log_2 nk^2)) 降成 (mathcal O(k^3+..)) 了。
#include <bits/stdc++.h>
const int N = 500005, K = 505, P = 1e9 + 7, inv2 = P + 1 >> 1;
int n, m, l, a[K], b[N], c[N], ans = 0; char str[N];
int inv[K], fac[K], ifac[K], pw[K], f0[K], f1[K], g0[K], g1[K], y[K];
int C(int n, int m) { return 1LL * fac[n] * ifac[m] % P * ifac[n - m] % P; }
int lagrange(int n, int k) {
for (int i = 0; i <= k + 1; i++) {
y[i] = 1;
for (int j = 1; j <= k; j++)
y[i] = 1LL * y[i] * i % P;
if (i) y[i] = (y[i] + y[i - 1]) % P;
}
int ans = 0;
for (int i = 0; i <= k + 1; i++) {
int tmp = y[i];
for (int j = 0; j <= k + 1; j++)
if (i != j) tmp = 1LL * tmp * (n - j + P) % P * (i > j ? inv[i - j] : P - inv[j - i]) % P;
ans = (ans + tmp) % P;
}
return ans;
}
int main() {
scanf("%s%d", str, &m);
for (int i = 0; i < m; i++) scanf("%d", &a[i]);
n = strlen(str); l = std::min(n, m);
std::reverse(str, str + n);
for (int i = n - 1; ~i; i--)
b[i] = (2 * b[i + 1] + (str[i] == '1')) % P, c[i] = c[i + 1] ^ (str[i] == '1');
inv[1] = 1;
for (int i = 2; i <= m; i++) inv[i] = 1LL * (P - P / i) * inv[P % i] % P;
fac[0] = ifac[0] = pw[0] = 1;
for (int i = 1; i <= m; i++)
fac[i] = 1LL * fac[i - 1] * i % P, ifac[i] = 1LL * ifac[i - 1] * inv[i] % P, pw[i] = 2 * pw[i - 1] % P;
f0[0] = 1;
for (int i = 0; i < l; i++) {
if (str[i] == '1') {
int x = 1LL * b[i + 1] * pw[i + 1] % P;
for (int j = i; j < m; j++)
for (int k = 0, t = a[j]; k <= j; k++)
ans = (ans + 1LL * C(j, k) * t % P * (c[i] ? f1[j - k] : f0[j - k])) % P, t = 1LL * t * x % P;
}
for (int j = 0; j < m; j++) {
g0[j] = g1[j] = 0;
for (int k = j; ~k; k--)
g0[j] = (g0[j] * 2 + 1LL * C(j, k) * f1[k]) % P,
g1[j] = (g1[j] * 2 + 1LL * C(j, k) * f0[k]) % P;
}
for (int j = 0; j < m; j++)
f0[j] = (1LL * f0[j] * pw[j] + g0[j]) % P, f1[j] = (1LL * f1[j] * pw[j] + g1[j]) % P;
}
for (int i = 0; i < m; i++)
ans = (ans + 1LL * a[i] * lagrange(1LL * b[i + 1] * pw[i + 1] % P - 1, i) % P * inv2) % P;
printf("%d", ans);
return 0;
}