#include <bits/stdc++.h>
using namespace std;
typedef long long lld;
const int MOD = 1e9 + 7;
const int MN = 105;
int fpow(int n, int k)
{
if(k == 0) return 1;
if(k % 2){
return 1ll * n * fpow(n, k - 1) % MOD;
}
int h = fpow(n, k / 2);
return 1ll * h * h % MOD;
}
int inv(int n)
{
return 1ll * fpow(n, MOD - 2) % MOD;
}
inline void mult(int& a, int b)
{
a = 1ll * a * b % MOD;
}
int mult2(int a, int b)
{
return 1ll * a * b % MOD;
}
inline void sub(int& a, int b)
{
a = ((a - b) % MOD + MOD) % MOD;
}
int rational(int a, int b)
{
return 1ll * a * inv(b) % MOD;
}
void add(int &a, int b)
{
a = ((a + b) % MOD + MOD) % MOD;
}
vector<vector<int>> inverse(const vector<vector<int>>& mat)
{
vector<vector<int>> I(mat.size(), vector<int>(mat.size(), 0));
vector<vector<int>> Q = mat;
int n = mat.size();
for(int i = 0; i < n; ++i) I[i][i] = 1;
for(int i = 0; i < n; ++i){
if(Q[i][i] == 0){
for(int j = i + 1; j < n; ++j){
if(Q[j][i] != 0){
for(int k = 0; k < n; ++k){
add(Q[i][k], Q[j][k]);
add(I[i][k], I[j][k]);
}
break;
}
}
}
int b = inv(Q[i][i]);
for(int j = 0; j < n; ++j){
mult(Q[i][j], b);
mult(I[i][j], b);
}
for(int j = 0; j < n; ++j){
if(i == j) continue;
int w = Q[j][i];
for(int k = 0; k < n; ++k){
sub(Q[j][k], mult2(Q[i][k], w));
sub(I[j][k], mult2(I[i][k], w));
}
}
}
return I;
}
vector<vector<int>> mult(const vector<vector<int>>& a, const vector<vector<int>>& b)
{
vector<vector<int>> ret(a.size(), vector<int>(b[0].size()));
for(int i = 0; i < a.size(); ++i){
for(int j = 0; j < b[0].size(); ++j){
int sum = 0;
for(int k = 0; k < a[0].size(); ++k){
add(sum, (1ll * a[i][k] * b[k][j]) % MOD);
}
ret[i][j] = sum;
}
}
return ret;
}
int main()
{
int p, q, n, k;
cin >> p >> q >> n >> k;
vector<vector<int>> Q(n - 1, vector<int>(n - 1, 0));
for(int i = 0; i + 1 < Q.size(); ++i){
Q[i][i+1] = rational(p - q,p);
Q[i+1][i] = rational(q, p);
}
for(int i = 0; i < Q.size(); ++i){
for(int j = 0; j < Q.size(); ++j){
Q[i][j] = ((i == j) - Q[i][j] + MOD) % MOD;
}
}
vector<vector<int>> N = inverse(Q);
vector<vector<int>> R(n - 1, vector<int>(2, 0));
R[0][0] = rational(q, p);
R[n - 2][1] = rational(p - q, p);
vector<vector<int>> B = mult(N, R);
cout << B[k - 1][1] << '\n';
}
# |
결과 |
실행 시간 |
메모리 |
Grader output |
1 |
Runtime error |
7 ms |
1720 KB |
SIGSEGV Segmentation fault |
2 |
Halted |
0 ms |
0 KB |
- |