Everyone will meet some difficult
Topic background:
Analysis: Mathematics + Number Theory
I don't know how to do 100 points, and I don't want to learn it, so let's talk about 80 points. First of all, we can know that the answer is
Obviously, for a fixed k , the number of combinations is a polynomial of degree m - n , and for any k , the coefficients of this polynomial are obviously the same, then we define that the coefficient of the i -th degree of k is a k , then obviously:
Obviously for a i , we can directly demolition the number of combinations in (m - n) 2 time.
So consider the latter part directly, let:
The above recursion can be optimized directly with the fast power of the matrix.
Time complexity O((m - n) 3 log(m- n)) .
Source:
/*
created by scarlyw
*/
#include <cstdio>
#include <string>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <cmath>
#include <cctype>
#include <vector>
#include <set>
#include <queue>
#include <ctime>
#include <bitset>
inline char read() {
static const int IN_LEN = 1024 * 1024;
static char buf[IN_LEN], *s, *t;
if (s == t) {
t = (s = buf) + fread(buf, 1, IN_LEN, stdin);
if (s == t) return -1;
}
return *s++;
}
/*
template<class T>
inline void R(T &x) {
static char c;
static bool iosig;
for (c = read(), iosig = false; !isdigit(c); c = read()) {
if (c == -1) return ;
if (c == '-') iosig = true;
}
for (x = 0; isdigit(c); c = read())
x = ((x << 2) + x << 1) + (c ^ '0');
if (iosig) x = -x;
}
//*/
const int OUT_LEN = 1024 * 1024;
char obuf[OUT_LEN], *oh = obuf;
inline void write_char(char c) {
if (oh == obuf + OUT_LEN) fwrite(obuf, 1, OUT_LEN, stdout), oh = obuf;
*oh++ = c;
}
template<class T>
inline void W(T x) {
static int buf[30], cnt;
if (x == 0) write_char('0');
else {
if (x < 0) write_char('-'), x = -x;
for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 + 48;
while (cnt) write_char(buf[cnt--]);
}
}
inline void flush() {
fwrite(obuf, 1, oh - obuf, stdout);
}
///*
template<class T>
inline void R(T &x) {
static char c;
static bool iosig;
for (c = getchar(), iosig = false; !isdigit(c); c = getchar())
if (c == '-') iosig = true;
for (x = 0; isdigit(c); c = getchar())
x = ((x << 2) + x << 1) + (c ^ '0');
if (iosig) x = -x;
}
//*/
const int MAXN = 1000000 + 10;
const int mod = 1000000000 + 7;
long long fac[MAXN], inv_fac[MAXN];
long long s, t, n, m;
inline long long mod_pow(long long a, long long b) {
int years = 1;
for (; b; b >>= 1, a = a * a % mod)
if (b & 1) ans = ans * a % mod;
return ans;
}
inline void get_c() {
fac [0] = 1;
for (int i = 1; i < MAXN; ++i) fac[i] = fac[i - 1] * i % mod;
inv_fac [MAXN - 1] = mod_pow (fac [MAXN - 1], mod - 2);
for (int i = MAXN - 2; i >= 0; --i)
inv_fac [i] = inv_fac [i + 1] * (i + 1)% mod;
}
inline long long c(int n, int m) {
if (n < m) return 0;
return fac [n] * inv_fac [m]% mod * inv_fac [n - m]% mod;
}
inline void solve_1() {
get_c();
long long ans = 0;
for (int i = 0, sign = 1; i <= n; ++i, sign = -sign) {
ans = (((ans + (long long)sign * c(n, i) * c(s - i * t, m))
% mod) + mod)% mod;
}
std::cout << ans;
}
const int MAXD = 100 + 10;
struct matrix {
int n;
long long a[MAXD][MAXD];
matrix(int n = 0) : n(n) {
for (int i = 0; i <= n; ++i)
for (int j = 0; j <= n; ++j)
a[i][j] = 0;
}
inline matrix operator * (const matrix &c) const {
matrix ret(n);
for (int i = 0; i <= n; ++i)
for (int k = 0; k <= n; ++k)
for (int j = 0; j <= n; ++j)
ret.a[i][j] += a[i][k] * c.a[k][j] % mod;
for (int i = 0; i <= n; ++i)
for (int j = 0; j <= n; ++j)
ret.a [i] [j]% = mod;
return ret;
}
inline matrix operator ^ (int b) const {
matrix ans(n), a = *this;
for (int i = 0; i <= n; ++i) ans.a[i][i] = 1;
for (; b; b >>= 1, a = a * a)
if (b & 1) ans = ans * a;
return ans;
}
} ;
long long last[MAXD], cur[MAXD], sum[MAXD], mul[MAXN];
inline void solve_2() {
long long ans = 0;
last[0] = 1, get_c(), s %= mod;
for (int i = 0; i < m - n; ++i) {
for (int j = 0; j <= i + 1; ++j) cur[j] = 0;
for (int j = 0; j <= i; ++j) cur[j + 1] = last[j];
for (int j = 0; j <= i; ++j)
cur[j] = (cur[j] + last[j] * (s - i) % mod) % mod;
for (int j = 0; j <= i + 1; ++j) last[j] = cur[j];
}
long long ret = 1;
for (int i = 1; i <= m - n; ++ i) ret = ret * i% mod;
ret = mod_pow (ret, mod - 2);
for (int i = 0; i <= m - n; ++i) cur[i] = cur[i] * ret % mod;
for (int i = 1; i <= t; ++i) mul[i] = 1;
sum[0] = t;
for (int i = 1; i <= m - n; ++i) {
for (int j = 1; j <= t; ++j)
mul[j] = mul[j] * j % mod, sum[i] += mul[j];
sum[i] %= mod;
}
matrix move(m - n);
for (int i = 0; i <= m - n; ++i)
for (int j = 0; j <= m - n; ++j)
move.a[i][j] = c(j, i) * sum[j - i] % mod;
move = (move ^ n);
// for (int i = 0; i <= m - n; ++i, std::cout << '\n')
// for (int j = 0; j <= m - n; ++j)
// std::cout << move.a[i][j] << " ";
for (int i = 0, sign = 1; i <= m - n; ++i, sign = -sign)
ans += (long long)sign * cur[i] * move.a[0][i] % mod;
ans = (ans% mod + mod)% mod;
std::cout << ans;
}
int main() {
freopen("success.in", "r", stdin);
freopen("success.out", "w", stdout);
R(s), R(t), R(n), R(m);
if (s < MAXN && m < MAXN) solve_1();
else solve_2();
return 0;
}