牛客多校训练2 B.Eddy Walker 2
原题地址：https://ac.nowcoder.com/acm/contest/882/B
题意：从 0 0 0点出发，给出一个 k k k，每次都有 1 k \frac{1}{k} k1的概率走 1 1 1步，有 1 k \frac{1}{k} k1的概率走 2 2 2步…有 1 k \frac{1}{k} k1的概率走 k k k步,给出 n n n，求走到 n n n点的概率。
思路：首先很容易的可以得到一个 d p dp dp转移式。
令 d p [ i ] = j dp[i]=j dp[i]=j,表示走到 i i i点的概率。那么 d p [ i ] = ∑ j = i − k i − 1 d p [ j ] k dp[i]=\frac{\sum_{j=i-k}^{i-1}dp[j]}{k} dp[i]=k∑j=i−ki−1dp[j]
这个式子只与前 k k k项有关，所以可以进行线性递推的优化。
注意需要特判再无穷远处的取值= 2 k + 1 \frac{2}{k+1} k+12
解释：我们可以计算出走一步的期望，也就是 1 k + 2 k + 3 k + . . . + k k = k + 1 2 \frac{1}{k}+\frac{2}{k}+\frac{3}{k}+...+\frac{k}{k}=\frac{k+1}{2} k1+k2+k3+...+kk=2k+1,那么走一步到达恰好到达一个点的概率就是 1 k + 1 2 = 2 k + 1 \frac{1}{\frac{k+1}{2}}=\frac{2}{k+1} 2k+11=k+12
用BM写的代码
#include
#define pb push_back
#define SZ(x) ((ll)(x).size())
typedef vector<ll> VI;
typedef pair<ll, ll> PII;
const ll mod = 1000000007;
ll powmod(ll a, ll b) {
ll res = 1;
a %= mod;
assert(b >= 0);
for (; b; b >>= 1) {
if (b & 1)res = res * a % mod;
a = a * a % mod;
}
return res;
}
// head
ll _;
namespace linear_seq {
const ll N = 10010;
ll res[N], base[N], _c[N], _md[N];
vector<ll> Md;
void mul(ll *a, ll *b, ll k) {
rep(i, 0, k + k) _c[i] = 0;
rep(i, 0, k) if (a[i]) rep(j, 0, k) _c[i + j] = (_c[i + j] + a[i] * b[j]) % mod;
for (ll i = k + k - 1; i >= k; i--) if (_c[i])
rep(j, 0, SZ(Md)) _c[i - k + Md[j]] = (_c[i - k + Md[j]] - _c[i] * _md[Md[j]]) % mod;
rep(i, 0, k) a[i] = _c[i];
}
ll solve(ll n, VI a, VI b) { // a 系数 b 初值 b[n+1]=a[0]*b[n]+...
// printf("%d\n",SZ(b));
ll ans = 0, pnt = 0;
ll k = SZ(a);
assert(SZ(a) == SZ(b));
rep(i, 0, k) _md[k - 1 - i] = -a[i];
_md[k] = 1;
Md.clear();
rep(i, 0, k) if (_md[i] != 0) Md.push_back(i);
rep(i, 0, k) res[i] = base[i] = 0;
res[0] = 1;
while ((1ll << pnt) <= n) pnt++;
for (ll p = pnt; p >= 0; p--) {
mul(res, res, k);
if ((n >> p) & 1) {
for (ll i = k - 1; i >= 0; i--) res[i + 1] = res[i];
res[0] = 0;
rep(j, 0, SZ(Md)) res[Md[j]] = (res[Md[j]] - res[k] * _md[Md[j]]) % mod;
}
}
rep(i, 0, k) ans = (ans + res[i] * b[i]) % mod;
if (ans < 0) ans += mod;
return ans;
}
VI BM(VI s) {
VI C(1, 1), B(1, 1);
ll L = 0, m = 1, b = 1;
rep(n, 0, SZ(s)) {
ll d = 0;
rep(i, 0, L + 1) d = (d + (ll)C[i] * s[n - i]) % mod;
if (d == 0) ++m;
else if (2 * L <= n) {
VI T = C;
ll c = mod - d * powmod(b, mod - 2) % mod;
while (SZ(C) < SZ(B) + m) C.pb(0);
rep(i, 0, SZ(B)) C[i + m] = (C[i + m] + c * B[i]) % mod;
L = n + 1 - L;
B = T;
b = d;
m = 1;
} else {
ll c = mod - d * powmod(b, mod - 2) % mod;
while (SZ(C) < SZ(B) + m) C.pb(0);
rep(i, 0, SZ(B)) C[i + m] = (C[i + m] + c * B[i]) % mod;
++m;
}
}
return C;
}
ll gao(VI a, ll n) {
VI c = BM(a);
c.erase(c.begin());
rep(i, 0, SZ(c)) c[i] = (mod - c[i]) % mod;
return solve(n, c, VI(a.begin(), a.begin() + SZ(c)));
}
};
int T;
ll k, n;
ll dp[5005];
int main() {
scanf("%d", &T);
while (T--) {
CLR(dp, 0);
scanf("%lld%lld", &k, &n);
if (n == 0) {
printf("1\n");
continue;
}
if (n == -1) {
printf("%lld\n", 2LL * powmod(k + 1, mod - 2) % mod);
continue;
}
vector<ll>v;
dp[0] = 1;
v.push_back(1);
for (int i = 1; i <= k; i++) {
for (int j = 0; j < i; j++) {
dp[i] = (dp[i] + dp[j]) % mod;
}
dp[i] = dp[i] * powmod(k, mod - 2) % mod;
v.push_back(dp[i]);
}
for (int i = k + 1; i <= 2 * k + 1; i++) {
for (int j = 1; j <= k; j++) {
dp[i] = (dp[i] + dp[i - j]) % mod;
}
dp[i] = dp[i] * powmod(k, mod - 2) % mod;
v.push_back(dp[i]);
}
printf("%lld\n", linear_seq::gao(v, n));
}
return 0;
}