题解 | #网格走法数目#

#include <iostream>
#include <vector>
using namespace std;

int uniquePaths(int x, int y) {
    // 创建动态规划数组
    vector<vector<int>> dp(x + 1, vector<int>(y + 1, 0));
    
    // 初始化第一行和第一列
    for (int i = 1; i <= x; i++) dp[i][1] = 1;
    for (int j = 1; j <= y; j++) dp[1][j] = 1;
    
    // 填充dp数组
    for (int i = 2; i <= x; i++) {
        for (int j = 2; j <= y; j++) {
            dp[i][j] = dp[i-1][j] + dp[i][j-1];
        }
    }
    
    return dp[x][y];
}

int main() {
    int x, y;
    while (cin >> x >> y) {
        cout << uniquePaths(x + 1, y + 1) << endl;
    }
    return 0;
}

import java.util.Scanner;

public class Main {
    public static int uniquePaths(int x, int y) {
        // 创建动态规划数组
        int[][] dp = new int[x + 1][y + 1];
        
        // 初始化第一行和第一列
        for (int i = 1; i <= x; i++) dp[i][1] = 1;
        for (int j = 1; j <= y; j++) dp[1][j] = 1;
        
        // 填充dp数组
        for (int i = 2; i <= x; i++) {
            for (int j = 2; j <= y; j++) {
                dp[i][j] = dp[i-1][j] + dp[i][j-1];
            }
        }
        
        return dp[x][y];
    }
    
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        while (sc.hasNext()) {
            int x = sc.nextInt();
            int y = sc.nextInt();
            System.out.println(uniquePaths(x, y));
        }
    }
}

def unique_paths(x: int, y: int) -> int:
    # 创建动态规划数组
    dp = [[0] * (y + 1) for _ in range(x + 1)]
    
    # 初始化第一行和第一列
    for i in range(1, x + 1):
        dp[i][1] = 1
    for j in range(1, y + 1):
        dp[1][j] = 1
        
    # 填充dp数组
    for i in range(2, x + 1):
        for j in range(2, y + 1):
            dp[i][j] = dp[i-1][j] + dp[i][j-1]
            
    return dp[x][y]

# ACM模式处理输入输出
while True:
    try:
        x, y = map(int, input().split())
        print(unique_paths(x + 1, y + 1))
    except EOFError:
        break