排列组合与杨辉三角类比运算

                                             1                                           1  1                                          1  2  1                                        1  3  3  1                                      1  4  6  4  1                                   

此杨辉三角计算公式: a[i][j] = a[i-1][j-1]+a[i-1][j];

与排列组合中:C​i​j​​   = C​i−1​j​​+C​i−1​j−1​​

具有类似之处，

来一道应用题,

假设要从一个 n 行 m 列的格子的左上角走到右下角，每次只能向右走一步，或者向下走一步，那么有多少种走法?

这道题其实也有动态的身影，

输入格式,(1=<n<=3,3<m<=1000000000)

#include <iostream>using namespace std;int main(){    long long n,m;    cin>>n>>m;    if(n == 1){        cout<<1<<endl;        return 0;    }    long long one = 1;    long long  two = 0;    long long three = 0;    if(n == 2)    for(int i = 1;i <= m;i ++){        two = two + one;    }    if(n == 3)    for(int i = 1;i <=m ;i ++){        two = two + one;        three = three + two;    }    if(n == 2)        cout<<two<<endl;    if(n == 3)        cout<<three<<endl;    return 0;}