AcWing 2067. 走方格
原题链接
简单
作者:
冉俊泽
,
2021-02-10 12:07:44
,
所有人可见
,
阅读 254
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递推的做法
#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
const int N = 50;
int a[N][N];
int n, m;
int main()
{
cin >> n >> m;
for(int i = 0; i < 33; i++)
{
a[i][0] = 1;
a[0][i] = 1;
}
for(int i = 1; i <= n; i++)
{
for(int j = 1; j <= m; j++)
{
if(i % 2 != 0 || j % 2 != 0)
{
if(i == 1)
a[i][j] = a[i][j - 1];
else if(j == 1)
a[i][j] = a[i - 1][j];
else
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
}
cout << a[n][m] << endl;
return 0;
}