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棋盘左上角到右下角方案数(2)
2022-07-06 12:18:00 【朴小明】
1.从左上角(1,1)到右下角(n,m)的方案数是C(n + m - 2, n - 1)
2.左上角到右下角的方案数之和是 C(n + m - 1, n)- 1
3.第n行的和是C(n + m - 1, n)
第三点的证明,可以看E题的题解Codeforces Global Round 21 A-E - kilo的文章 - 知乎
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