大佬求解(^v^)

#include <iostream>
#include <vector>
using namespace std;
const int MOD = 1e9 + 7;
int countEscapePlans(vector<vector<int>>& gates) {
    int N = gates.size();
    int K = gates[0].size();
    vector<vector<int>> dp(N, vector<int>(K, 0));
    // 初始化边界条件
    dp[0][0] = 1;
    // 动态规划求解
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < K; j++) {
            if (i > 0) {
                dp[i][j] += dp[i - 1][j];
                dp[i][j] %= MOD;
            }
            if (j > 0) {
                dp[i][j] += dp[i][j - 1];
                dp[i][j] %= MOD;
            }
        }
    }
    return dp[N - 1][K - 1];
}
int main() {
    int N, K;
    cin >> N >> K;
    vector<vector<int>> gates(N, vector<int>(K, 0));
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < K; j++) {
            cin >> gates[i][j];
        }
    }
    int result = countEscapePlans(gates);
    cout << result << endl;
    return 0;
}