题解 | #输出单向链表中倒数第k个结点#

#include<stdio.h>


int main()
{
        //动态规划
    //dp[m][n]:m个苹果放到n个盘子***有dp[m][n]中分法
    //(1)如果0个苹果或者1个苹果，或者1个盘子：dp[0][n]==dp[1][n]==dp[m][1]==1；
    //(2)如果有盘子为空，dp[m][n] = dp[m][n-1];
    //(3)如果没有盘子为空，dp[m][n] = dp[m-n][n];
    int m, n;
    int dp[11][11] = { 0 };
    while (scanf("%d%d", &m, &n) != EOF)
    {
        for (int i = 0; i <= m; i++)
        {
            for (int j = 1; j <= n; j++)
            {
                if (i == 0 || i == 1 || j == 1)
                {
                    dp[i][j] = 1;
                }
                else if (i >= j)
                {
                    dp[i][j] = dp[i - j][j] + dp[i][j - 1];
                }
                else
                {
                    dp[i][j] = dp[i][j - 1];
                }
            }
        }
        printf("%d\n", dp[m][n]);
    }

}