t3考虑容斥，代码如下： mod = 10 ** 9 + 7 N = 2 * 10 ** 6 + 10 fac = [1] * N for i in range(2, N): fac[i] = fac[i - 1] * i % mod invfac = [1] * N invfac[N - 1] = pow(fac[N - 1], mod - 2, mod) for i in range(N - 1)[::-1]: invfac[i] = invfac[i + 1] * (i + 1) % mod def c(i, j): return fac[i] * invfac[j] * invfac[i - j] % mod n = 1000000 ans = 0 mx = n // 3 for i in range(2, mx + 1): if i % 2 == 0: ans += (i - 1) * c(i + n - 3 * i, i) * pow(26, n - 3 * i, mod) % mod else: ans -= (i - 1) * c(i + n - 3 * i, i) * pow(26, n - 3 * i, mod) % mod ans %= mod print(ans)