穿过街道(递推+记忆化搜索俩种写法)

穿过街道

Problem:20

Time Limit:1000ms

Memory Limit:65536K

Description

一个城市的街道布局如下:从最左下方走到最右上方,每次只能往上或往右走,一共有多少种走法? 

Input

输入很多行行数,每行1个数字代表n的值,当n=0时结束(2<=n<=15)

Output

输出对应每行n值的走法.

Sample Input

1
2
10
5
0

Sample Output

2
6
184756
252

dp入门题，当前这一点的方法就等于我来自于左面和下面的和。

记忆化搜索写法:

# include 
# include 

using namespace std;

const int maxn = 20;

int dp[maxn][maxn];

int dfs(int x, int y)
{
    if(dp[x][y] != -1)
    {
        return dp[x][y];
    }
    int result;
    if(x == 0|| y== 0)
    {
        return 1;
    }
    else
    {
        dp[x][y] = dfs(x - 1, y) + dfs(x, y - 1);
        result = dp[x][y];
        return result;
    }
}

int main(int argc, char *argv[])
{
    int n;
    memset(dp, -1, sizeof(dp));
    while(cin >> n, n)
    {

        cout << dfs(n, n) << endl;
    }
    return 0;
}

 

 

递推写法:

# include 
# include 

using namespace std;

const int maxn = 20;

int dp[maxn][maxn];

int main(int argc, char *argv[])
{
    int n;
    memset(dp, 0, sizeof(dp));
    for(int i = 0; i < maxn; i++)
    {
        dp[i][0] = 1;
        dp[0][i] = 1;
    }
    for(int i = 1; i < maxn; i++)
    {
        for(int j = 1; j < maxn; j++)
        {
            dp[i][j] = dp[i][j - 1] + dp[i - 1][j];
        }
    }
    while(cin >> n, n)
    {
        cout << dp[n][n] << endl;
    }
    return 0;
}