計蒜客之動態規劃篇(1)

1. 爬樓梯ios //用組合數的方法比較拖沓 #include <iostream> using namespace std; const int N = 55; int n; unsigned dp[N]; //dp[i]:從階梯0跳到n的方法數, 爆int void solve() { dp[1] = dp[0] = 1; for (int i = 2; i <= n;

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