嘟嘟嘟
题目有点坑，要你求的多少大阵指的是召唤kkk的大阵数 * lzn的大阵数，不是相加。
看到这个限制条件，显然要用生成函数推一推。
比如第一个条件“金神石的块数必须是6的倍数”，就是\(1 +x ^ 6 + x ^ {12} + \ldots\)，也就是\(\frac{1 - x ^ {6n}}{1 - x ^ 6}\)。当\(x \in (-1, 1)\)时，就变成了\(\frac{1}{1 - x ^ 6}\)。
剩下的同理。
然后把这10个条件都乘起来，一顿化简，答案就是\(\frac{(n + 1) * (n + 2) * (n + 3) *(n + 4)}{24}\)。
本来想快乐的写高精，但是\(n = 1e5\)还非得用fft。
于是就写了一发，不开O2会TLE飞，开了后TLE最后一个点。然后把fft的预处理改成bin哥的写法后就过了。
#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const db PI = acos(-1);
const int maxn = 4e6 + 5;
inline ll read()
{
ll ans = 0;
char ch = getchar(), last = ' ';
while(!isdigit(ch)) last = ch, ch = getchar();
while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
if(last == '-') ans = -ans;
return ans;
}
inline void write(ll x)
{
if(x < 0) x = -x, putchar('-');
if(x >= 10) write(x / 10);
putchar(x % 10 + '0');
}
char a1[maxn];
int n, m, a[maxn], b[maxn];
int len = 1;
struct Comp
{
db x, y;
In Comp operator + (const Comp& oth)const
{
return (Comp){x + oth.x, y + oth.y};
}
In Comp operator - (const Comp& oth)const
{
return (Comp){x - oth.x, y - oth.y};
}
In Comp operator * (const Comp& oth)const
{
return (Comp){x * oth.x - y * oth.y, x * oth.y + y * oth.x};
}
friend In void swap(Comp& a, Comp& b)
{
swap(a.x, b.x); swap(a.y, b.y);
}
}c[maxn], d[maxn], omg[maxn], inv[maxn];
int r[maxn];
In void init()
{
omg[0] = inv[0] = (Comp){1, 0};
omg[1] = inv[len - 1] = (Comp){cos(2 * PI / len), sin(2 * PI / len)};
for(int i = 2; i < len; ++i) omg[i] = inv[len - i] = omg[i - 1] * omg[1];
}
In void fft(Comp* a, Comp* omg)
{
for(int i = 0; i < len; ++i) if(i < r[i]) swap(a[i], a[r[i]]);
for(int l = 2; l <= len; l <<= 1)
{
int q = l >> 1;
for(Comp* p = a; p != a + len; p += l)
for(int i = 0; i < q; ++i)
{
Comp tp = omg[len / l * i] * p[i + q];
p[i + q] = p[i] - tp, p[i] = p[i] + tp;
}
}
}
In void mul(int* a, int* b)
{
int tot = max(n, m); len = 1;
while(len < (tot << 1)) len <<= 1;
int lim = 0;
while((1 << lim) < len) ++lim;
for(int i = 0; i < len; ++i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (lim - 1));
for(int i = 0; i < len; ++i) c[i] = d[i] = (Comp){0, 0};
for(int i = 0; i < n; ++i) c[i] = (Comp){a[i], 0};
for(int i = 0; i < m; ++i) d[i] = (Comp){b[i], 0};
init();
fft(c, omg), fft(d, omg);
for(int i = 0; i < len; ++i) c[i] = c[i] * d[i];
fft(c, inv);
for(int i = 0; i <= len; ++i) a[i] = 0;
for(int i = 0; i < len; ++i)
{
a[i] += (int)(c[i].x / len + 0.5);
if(a[i] >= 10) a[i + 1] += a[i] / 10, a[i] %= 10;
}
n = len;
while(n - 1 && !a[n - 1]) --n;
// for(int i = n - 1; i >= 0; --i) printf("%d", a[i]); enter;
}
In void add(int* a, int x, int& n)
{
a[0] += x;
for(int i = 0; i < n; ++i)
if(a[i] >= 10) a[i + 1] += a[i] / 10, a[i] %= 10;
else break;
++n;
while(n - 1 && !a[n - 1]) --n;
// for(int i = n - 1; i >= 0; --i) printf("%d", a[i]); enter;
}
In void div(int* a, int x)
{
static int ret[maxn];
reverse(a, a + n);
int tp = 0, cnt = 0;
for(int i = 0; i < n; ++i)
{
tp = tp * 10 + a[i];
ret[++cnt] = tp / x;
tp %= x;
}
int sta = 1;
while(sta < cnt && !ret[sta]) ++sta;
for(int i = sta; i <= cnt; ++i) write(ret[i]); enter;
}
int main()
{
// freopen("random.in", "r", stdin);
// freopen("ac.out", "w", stdout);
scanf("%s", a1);
m = n = strlen(a1);
for(int i = 0; i < n; ++i) b[i] = a[i] = a1[n - i - 1] - '0';
add(a, 1, n); add(b, 1, m);
for(int i = 2; i <= 4; ++i)
{
add(b, 1, m);
mul(a, b);
}
div(a, 24);
return 0;
}