2021CCPC威海 M.810975
作者：互联网
题意：问构造出长度为 \(n\) 的01串，有 \(m\) 个1，其中最长连续 \(1\) 的段长度恰好为 \(k\) 的方案数。
知识点：容斥，多项式快速幂
先推荐一个类似的题目 HDU6397 Character Encoding
这题有两种方法，先说简单的那种
可以先解决将 \(m\) 个 \(1\) 插入到 \(n - m + 1\) 个空中，其中最长连续 \(1\) 的段长度不超过 \(k\) 的方案数。
这个问题等价与问
\[\sum_{i=1}^{n-m+1}x_i = m(0 \le x_i \le k) \]方程组非负整数解的个数
于是对于每个 \(x_i\) 有生成函数 \(\sum_{i=0}^k x^i\)
得到答案多项式为
\[G(x) = (\sum_{i=0}^k x^i)^{n-m+1} \]这个可以用多项式快速幂求出
\([x^m]G(x)\) 即是答案
当然这个是最长段不超过 \(k\) 的答案，只需要减去最长段不超过 \(k-1\) 的答案即可
注意特判即可
#include <bits/stdc++.h>
#define endl '\n'
#define ls u << 1
#define rs u << 1 | 1
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL,LL> PLL;
const int INF = 0x3f3f3f3f, N = 1e5 + 10;
const double eps = 1e-6;
const double PI = acos(-1);
inline int lowbit(int x) {return x & (-x);}
namespace polybase {//范围为1e9需要先取模,别忘记改模数,原根和数组长度
constexpr LL MOD = 998244353, G = 3, L = 1 << 22;
//分别表示模数，原根以及默认数组长度
LL fpow(LL x, LL r) {
LL result = 1;
while (r) {
if (r & 1) result = result * x % MOD;
r >>= 1;
x = x * x % MOD;
}
return result;
}
int w[L], _ = [] {
LL x = fpow(G, (MOD - 1) / L);
w[L / 2] = 1;
for (int i = L / 2 + 1; i < L; i++)
w[i] = 1ll * w[i - 1] * x % MOD;
for (int i = L / 2 - 1; i >= 0; i--)
w[i] = w[i << 1];
return 0;
}();
inline int norm(int n) { return 1 << __lg(n * 2 - 1); }
void dft(LL *a, int n) {
for (int k = n >> 1; k; k >>= 1)
for (int i = 0; i < n; i += k << 1)
for (int j = 0; j < k; j++) {
LL &x = a[i + j], y = a[i + j + k];
a[i + j + k] = (x - y + MOD) * w[k + j] % MOD;
x += y;
if (x >= MOD)x -= MOD;
}
}
void idft(LL *a, int n) {
for (int k = 1; k < n; k <<= 1)
for (int i = 0; i < n; i += k << 1)
for (int j = 0; j < k; j++) {
LL x = a[i + j], y = a[i + j + k] * w[k + j] % MOD;
a[i + j + k] = x - y < 0 ? x - y + MOD : x - y;
a[i + j] += y;
if (a[i + j] >= MOD)a[i + j] -= MOD;
}
for (int i = 0, inv = MOD - (MOD - 1) / n; i < n; i++)
a[i] = a[i] * inv % MOD;
reverse(a + 1, a + n);
}
struct poly : public vector<LL> {
using vector<LL>::vector;
#define T (*this)
poly MODxk(int k) const {
k = min(k, (int) size());
return poly(begin(), begin() + k);
}
poly rev() const { return poly(rbegin(), rend()); }
friend void dft(poly &a) { dft(a.data(), a.size()); }
friend void idft(poly &a) { idft(a.data(), a.size()); }
friend poly operator*(const poly &x, const poly &y) {
if (x.empty() || y.empty())return poly();
poly a(x), b(y);
int len = a.size() + b.size() - 1;
int n = norm(len);
a.resize(n), b.resize(n);
dft(a), dft(b);
for (int i = 0; i < n; i++)
a[i] = a[i] * b[i] % MOD;
idft(a);
a.resize(len);
return a;
}
poly operator+(const poly &b) {
poly a(T);
if (a.size() < b.size())
a.resize(b.size());
for (int i = 0; i < b.size(); i++)//用b.size()防止越界
{
a[i] += b[i];
if (a[i] >= MOD)a[i] -= MOD;
}
return a;
}
poly operator-(const poly &b) {
poly a(T);
if (a.size() < b.size())
a.resize(b.size());
for (int i = 0; i < b.size(); i++) {
a[i] -= b[i];
if (a[i] < 0)a[i] += MOD;
}
return a;
}
poly operator*(const LL p) {
poly a(T);
for (auto &x: a)
x = x * p % MOD;
return a;
}
poly &operator<<=(int r) { return insert(begin(), r, 0), T; }//注意逗号,F(x)*(x^r)
poly operator<<(int r) const { return poly(T) <<= r; }
poly operator>>(int r) const { return r >= size() ? poly() : poly(begin() + r, end()); }
poly &operator>>=(int r) { return T = T >> r; }//F[x]/(x^r)
poly deriv() {
//求导
if (empty())return T;
poly a(size() - 1);
for (int i = 1; i < size(); i++)//注意是size()
a[i - 1] = T[i] * i % MOD;
return a;
}
poly integ() {
//积分
poly a(size() + 1);
for (int i = 1; i < a.size(); i++)//注意是a.size()
a[i] = T[i - 1] * fpow(i, MOD - 2) % MOD;
return a;
}
poly inv(int n) {
poly a{fpow(T[0], MOD - 2)};
int k = 1;
while (k < n) {
k <<= 1;
a = (a * 2 - MODxk(k) * a * a).MODxk(k);
}
return a.MODxk(n);
}
poly sqrt(int n) {
//f[0]必须等于1
poly a{1};
int k = 1;
const LL inv2 = fpow(2, MOD - 2);
while (k < n) {
k <<= 1;
a = ((MODxk(k) * a.inv(k)).MODxk(k) + a) * inv2;
}
return a.MODxk(n);
}
poly ln(int n) {
//需要保证f[0]=1
return (deriv() * inv(n)).integ().MODxk(n);
}
poly exp(int n) {
//需要保证f[0]=0
poly a{1};
int k = 1;
while (k < n) {
k <<= 1;
a = (a * (poly{1} - a.ln(k) + MODxk(k))).MODxk(k);
}
return a.MODxk(n);
}
#undef T
};
}
using namespace polybase;
inline void solve() {
int n, m, k; cin >> n >> m >> k;
if (m > n || k > m) {cout << 0 << endl; return ;}
if (m == 0) {cout << 1 << endl; return ;}
if (n == 0) {cout << 1 << endl; return ;}
if (k == 0) {cout << 0 << endl; return ;}
poly a(k + 1);
for (int i = 0; i <= k; i ++ ) a[i] = 1;
a = (a.ln(m + 1) * (n - m + 1)).exp(m + 1);
if (k == 1) {cout << a[m] << endl; return ;}
poly b(k);
for (int i = 0; i < k; i ++ ) b[i] = 1;
b = (b.ln(m + 1) * (n - m + 1)).exp(m + 1);
cout << (a[m] - b[m] + MOD) % MOD << endl;
}
signed main() {
#ifdef DEBUG
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
auto now = clock();
#endif
ios::sync_with_stdio(false), cin.tie(nullptr);
cout << fixed << setprecision(2);
// int T; cin >> T;
// while (T -- )
solve();
#ifdef DEBUG
cout << "============================" << endl;
cout << "Program run for " << (clock() - now) / (double)CLOCKS_PER_SEC * 1000 << " ms." << endl;
#endif
return 0;
}
接下来说另外一种方法
看回这个式子
\[\sum_{i=1}^{n-m+1}x_i = m(0 \le x_i \le k) \]先考虑没有 \(k\) 的限制，问题等价于 \(\sum_{i=1}^{n-m+1} x_i = m + k\) 的正整数解个数
可以隔板法得到答案 \(m + k - 1 \choose n-m+1\)
考虑容斥原理，分割成这样一个问题
设 $ans_k $ 为 \(\sum_{i=1}^{n-m+1} x_i = m (0 \le x_i \le k)\) 解的个数
这个解的方法可以参考 HDU6397 Character Encoding
则答案为 \(ans_k - ans_{k-1}\)
注意特判
#include <bits/stdc++.h>
#define endl '\n'
#define ls u << 1
#define rs u << 1 | 1
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL,LL> PLL;
const int INF = 0x3f3f3f3f, N = 3e5 + 10;
const int MOD = 998244353;
const double eps = 1e-6;
const double PI = acos(-1);
inline int lowbit(int x) {return x & (-x);}
LL f1[N], f2[N];
LL q_pow(LL a, LL b, LL p) {
LL res = 1;
for (; b; b >>= 1) {
if (b & 1) res = res * a % p;
a = a * a % p;
}
return res;
}
LL C(LL n, LL m) {
if (n < m) return 0;
if (n < 0 || m < 0) return 0;
if (n == m) return 1;
return f1[n] * f2[m] % MOD * f2[n - m] % MOD;
}
LL calc(LL n, LL m, LL k) {
LL res = 0;
int ops = 1;
for (LL i = 0; i <= m; i ++ ) {
res = (res + 1ll * ops * C(m, i) * C(m + k - 1 - i * n, m - 1) % MOD + MOD) % MOD;
ops *= -1;
}
return res;
}
inline void solve() {
LL n, m, k; cin >> n >> m >> k;
if (m > n || k > m) {cout << 0 << endl; return ;}
if (m == 0) {cout << 1 << endl; return ;}
if (n == 0) {cout << 1 << endl; return ;}
if (k == 0) {cout << 0 << endl; return ;}
cout << (calc(k + 1, n - m + 1, m) - calc(k, n - m + 1, m) + MOD) % MOD << endl;
}
signed main() {
#ifdef DEBUG
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
auto now = clock();
#endif
ios::sync_with_stdio(false), cin.tie(nullptr);
cout << fixed << setprecision(2);
int n = 300000;
f1[0] = 1;
for (int i = 1; i <= n; i ++ ) f1[i] = f1[i - 1] * i % MOD;
f2[n] = q_pow(f1[n], MOD - 2, MOD);
for (int i = n; i; i -- ) f2[i - 1] = f2[i] * i % MOD;
// int T; cin >> T;
// while (T -- )
solve();
#ifdef DEBUG
cout << "============================" << endl;
cout << "Program run for " << (clock() - now) / (double)CLOCKS_PER_SEC * 1000 << " ms." << endl;
#endif
return 0;
}
标签：return,int,LL,poly,2021CCPC,M.810975,威海,size,MOD 来源： https://www.cnblogs.com/JYF-AC/p/16559122.html