题解
小迪的blog ： https://www.cnblogs.com/RabbitHu/p/9178645.html
请大家点推荐并在sigongzi的评论下面点支持谢谢！
掌握了小迪生成函数的有趣姿势之后，我们考虑一下这个问题
由于出题人语死早，我们认为是十种石头的生成函数直接乘起来
(frac{1}{1 - x^6} cdot frac{1 - x^{10}}{1 - x} cdot frac{1 - x^{5}}{1 - x} cdot frac{1}{1 - x^4} cdot frac{1 - x^8}{1 - x} cdot frac{1}{1 - x^2} cdot frac{1 - x^2}{1 - x} cdot frac{1}{1 - x^8} cdot frac{1}{1 - x^10} cdot frac{1 - x^4}{1 - x} = frac{1}{(1 - x)^5})
这个正好是上面那个blog里和组合数有关的生成函数，所以我们要求的值就是
(inom{n + 5 - 1}{5 - 1} = inom{n + 4}{4} = frac{(n + 1)(n + 2)(n + 3)(n + 4)}{24})
显然需要高精度了，还得是FFT
可以用NTT代替（因为值不会太大）
“我司嘉祺就算是TLE死，死外面，从这里跳下去，也绝对不会开O2的！”
“氧气真好吸。”
代码
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <vector>
#define enter putchar('
')
#define space putchar(' ')
#define MAXN 1000005
//#define ivorysi
#define pb push_back
using namespace std;
typedef long long int64;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 - '0' + c;
c = getchar();
}
res = res * f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) out(x / 10);
putchar('0' + x % 10);
}
const int MOD = 998244353,MaxL = (1 << 20);
int W[(1 << 20) + 5],N;
char a[1000005];
int inc(int a,int b) {
return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
return 1LL * a * b % MOD;
}
int fpow(int x,int c) {
int res = 1,t = x;
while(c) {
if(c & 1) res = mul(res,t);
t = mul(t,t);
c >>= 1;
}
return res;
}
struct Bignum {
vector<int> v;
friend Bignum operator + (const Bignum &a,const int &x) {
Bignum c;c.v.clear();
int g = (a.v[0] + x) / 10;
c.v.pb((a.v[0] + x) % 10);
for(int i = 1 ; i < a.v.size() ; ++i) {
int t = a.v[i] + g;
c.v.pb(t % 10);g = t / 10;
}
if(g) c.v.pb(g);
return c;
}
friend void NTT(Bignum &f,int LEN,int on) {
f.v.resize(LEN);
for(int i = 1 , j = LEN / 2 ; i < LEN - 1 ; ++i) {
if(i < j) swap(f.v[i],f.v[j]);
int k = LEN / 2;
while(j >= k) {
j -= k;
k >>= 1;
}
j += k;
}
for(int h = 2 ; h <= LEN ; h <<= 1) {
int wn = W[(MaxL + on * MaxL / h) % MaxL];
for(int k = 0 ; k < LEN ; k += h) {
int w = 1;
for(int j = k ; j < k + h / 2 ; ++j) {
int u = f.v[j],t = mul(w,f.v[j + h / 2]);
f.v[j] = inc(u,t);
f.v[j + h / 2] = inc(u,MOD - t);
w = mul(w,wn);
}
}
}
if(on == -1) {
int InvL = fpow(LEN,MOD - 2);
for(int i = 0 ; i < LEN ; ++i) f.v[i] = mul(f.v[i],InvL);
}
}
friend Bignum operator * (Bignum a,Bignum b) {
int s = a.v.size() + b.v.size() - 2,t = 1;
while(t <= s) t <<= 1;
NTT(a,t,1);NTT(b,t,1);
Bignum c;c.v.clear();
for(int i = 0 ; i < t ; ++i) c.v.pb(mul(a.v[i],b.v[i]));
NTT(c,t,-1);
int64 x = 0;
for(int i = 0 ; i < t ; ++i) {
x += c.v[i];
c.v[i] = x % 10;
x /= 10;
}
while(x) {
c.v.pb(x % 10);
x /= 10;
}
for(int i = c.v.size() - 1 ; i > 0 ; --i) {
if(c.v[i] == 0) c.v.pop_back();
else break;
}
return c;
}
friend Bignum operator / (const Bignum &a,int k) {
Bignum c;c.v.clear();c.v.resize(a.v.size());
int x = 0;
for(int i = a.v.size() - 1 ; i >= 0 ; --i) {
x = x * 10 + a.v[i];
c.v[i] = x / k;
x %= k;
}
for(int i = a.v.size() - 1 ; i > 0 ; --i) {
if(c.v[i] == 0) c.v.pop_back();
else break;
}
return c;
}
void print() {
for(int i = v.size() - 1 ; i >= 0 ; --i) {
putchar('0' + v[i]);
}
}
}A,B;
void Solve() {
W[0] = 1;W[1] = fpow(3,(MOD - 1) / (1 << 20));
for(int i = 2 ; i < (1 << 20) ; ++i) W[i] = mul(W[i - 1],W[1]);
scanf("%s",a + 1);
N = strlen(a + 1);
A.v.clear();
for(int i = N ; i >= 1 ; --i) A.v.pb(a[i] - '0');
B = ((A + 1) * (A + 2)) * ((A + 3) * (A + 4));
B = B / 24;
B.print();enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}