https://leetcode.com/problems/2-keys-keyboard/description/
非常好的medium的题
这里有个证明 https://leetcode.com/problems/2-keys-keyboard/discuss/105900/C++-O(sqrt(n))-DP-and-greedy-prime-number-solution
里面a + b <= ab 稍微解释下:
1/b + 1/a <=1 而 a>=2 b>=2 所以成立
class Solution { public: int minSteps(int n) { if (n == 1) return 0; vector<int> dp(n+1, 0); // dp[1] = 1; dp[2] = 2; return dfs(dp, n); } int dfs(vector<int>&dp, int x) { if (dp[x] ) return dp[x]; int ans = x; for (int i = 2; i <= sqrt(x); i++) { if (x % i == 0) { ans = min( ans, dfs(dp, x/i) + i ); ans = min( ans, dfs(dp, i) + x/i ); } } dp[x] = ans; return dp[x]; } };
更好的还是
class Solution { public: int minSteps(int n) { int ans = 0; for (int i = 2; i <= int(sqrt(n)); i++) { while (n % i == 0) { n /= i; ans += i; } } if (n > 1) { ans += n; } return ans; } };