标签:函数 整数 vector ack sdi 西游 无向图 inline 有一个
为了减小文件大小,这里不写一堆题目背景了。
请写一个程序,输入一个数字N,输出N个点的森林的数量。点有标号。
森林是一种无向图,要求图中不能存在环(图可以不连通),或者说是由若干个树组成的集合。说到森林,我就想起今年下半年,中美合拍的西游记即将正式开机,我继续扮演美猴王孙悟空,我会用美猴王艺术形象努力创造一个正能量的形象,文体两开花,弘扬中华文化,希望大家能多多关注。
输入文件只有一个整数N。
输出森林的方案数。由于答案很大,所以请输出对998244353取模后的结果。
3
7
对于30%的数据满足3<=n<=9。
对于40%的数据满足3<=n<=90。
对于50%的数据3<=n<=900。
对于60%的数据满足3<=n<=9000。
对于100%的数据满足3<=n<=90000。
#include<bits/stdc++.h>
#define LL long long
LL in() {
char ch; LL x = 0, f = 1;
while(!isdigit(ch = getchar()))(ch == '-') && (f = -f);
for(x = ch ^ 48; isdigit(ch = getchar()); x = (x << 1) + (x << 3) + (ch ^ 48));
return x * f;
}
const int maxn = 4e6 + 100;
const int mod = 998244353;
using std::vector;
int len, r[maxn], fac[maxn];
LL ksm(LL x, LL y) {
LL re = 1LL;
while(y) {
if(y & 1) re = re * x % mod;
x = x * x % mod;
y >>= 1;
}
return re;
}
void FNTT(vector<int> &A, int flag) {
A.resize(len);
for(int i = 0; i < len; i++) if(i < r[i]) std::swap(A[i], A[r[i]]);
for(int l = 1; l < len; l <<= 1) {
int w0 = ksm(3, (mod - 1) / (l << 1));
for(int i = 0; i < len; i += (l << 1)) {
int a0 = i, a1 = i + l, w = 1;
for(int k = 0; k < l; k++, a0++, a1++, w = 1LL * w0 * w % mod) {
int tmp = 1LL * A[a1] * w % mod;
A[a1] = ((A[a0] - tmp) % mod + mod) % mod;
A[a0] = (A[a0] + tmp) % mod;
}
}
}
if(flag == -1) {
std::reverse(A.begin() + 1, A.end());
int inv = ksm(len, mod - 2);
for(int i = 0; i < len; i++) A[i] = 1LL * A[i] * inv % mod;
}
}
vector<int> operator * (vector<int> A, vector<int> B) {
int tot = A.size() + B.size() - 1;
for(len = 1; len <= tot; len <<= 1);
for(int i = 0; i < len; i++) r[i] = (r[i >> 1] >> 1) | ((i & 1) * (len >> 1));
vector<int> ans;
FNTT(A, 1), FNTT(B, 1);
for(int i = 0; i < len; i++) ans.push_back(1LL * A[i] * B[i] % mod);
FNTT(ans, -1);
ans.resize(tot);
return ans;
}
vector<int> operator - (const vector<int> &A, const vector<int> &B) {
vector<int> ans;
for(int i = 0; i < (int)std::min(A.size(), B.size()); i++) ans.push_back(A[i] - B[i]);
for(int i = A.size(); i < (int)B.size(); i++) ans.push_back(-B[i]);
for(int i = B.size(); i < (int)A.size(); i++) ans.push_back(A[i]);
return ans;
}
vector<int> operator + (const vector<int> &A, const vector<int> &B) {
vector<int> ans;
for(int i = 0; i < (int)std::min(A.size(), B.size()); i++) ans.push_back(A[i] + B[i]);
for(int i = A.size(); i < (int)B.size(); i++) ans.push_back(B[i]);
for(int i = B.size(); i < (int)A.size(); i++) ans.push_back(A[i]);
return ans;
}
vector<int> inv(const vector<int> &A) {
if(A.size() == 1) {
vector<int> ans;
ans.push_back(ksm(A[0], mod - 2));
return ans;
}
int n = A.size(), _ = (n + 1) >> 1;
vector<int> B = A, ans;
ans.push_back(2);
B.resize(_);
B = inv(B);
ans = B * (ans - A * B);
ans.resize(n);
return ans;
}
vector<int> getd(const vector<int> &A) {
vector<int> ans;
ans.resize(A.size() - 1);
for(int i = 1; i < (int)A.size(); i++) ans[i - 1] = 1LL * A[i] * i % mod;
return ans;
}
vector<int> geti(const vector<int> &A) {
vector<int> ans;
ans.resize(A.size() + 1);
for(int i = 0; i < (int)A.size(); i++) ans[i + 1] = 1LL * A[i] * ksm(i + 1, mod - 2) % mod;
return ans;
}
vector<int> getln(const vector<int> &A) { return geti(getd(A) * inv(A)); }
vector<int> getexp(const vector<int> &A) {
if(A.size() == 1) {
vector<int> ans;
ans.push_back(1);
return ans;
}
int n = A.size(), _ = (n + 1) >> 1;
vector<int> B = A, ans;
ans.push_back(1);
B.resize(_);
B = getexp(B);
ans = B * (ans - getln(B) + A);
ans.resize(n);
return ans;
}
int main() {
freopen("forest.in", "r", stdin);
freopen("forest.out", "w", stdout);
int n = in() + 1;
vector<int> a;
fac[0] = 1;
for(LL i = 1; i <= n; i++) fac[i] = 1LL * fac[i - 1] * i % mod;
a.push_back(0);
a.push_back(1);
for(int i = 2; i <= n; i++) a.push_back(1LL * ksm(i, i - 2) * ksm(fac[i], mod - 2) % mod);
a.resize(a.size() << 1);
a = getexp(a);
printf("%lld", 1LL * a[n - 1] * fac[n - 1] % mod);
return 0;
}
标签:函数 整数 vector ack sdi 西游 无向图 inline 有一个
原文地址:https://www.cnblogs.com/olinr/p/10359997.html