Submission #19486

#	Time	Username	Problem	Language	Result	Execution time	Memory
19486		algoshipda	Ω (kriii4_P3)	C++14	0 / 100	7 ms	1720 KiB

P3

#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';
}

• Subtask 1: 0 / 100