T(n) = 25T(n/5)+n^2的时间复杂度？

最新推荐文章于 2025-05-14 19:21:18 发布

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最新推荐文章于 2025-05-14 19:21:18 发布

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本文详细解析了递归公式T(n) = 25T(n/5) + n^2的时间复杂度，并通过主定理得出其为O(n^2*logn)。文中逐步展开递归过程，帮助理解递归函数的行为。

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原文地址：T(n) = 25T(n/5)+n^2的时间复杂度作者：djkpengjun
主定理：
对于T(n) = a*T(n/b)+c*n^k;T(1) = c 这样的递归关系，有这样的结论：
if (a > b^k) T(n) = O(n^(logb(a)));logb(a)b为底a的对数
if (a = b^k) T(n) = O(n^k*logn);
if (a < b^k) T(n) = O(n^k);
a=25; b = 5 ; k=2
a==b^k 故T(n)=O(n^k*logn)=O(n^2*logn)
T(n) = 25T(n/5)+n^2
= 25(25T(n/25)+n^2/25)+n^2
= 625T(n/25)+n^2+n^2 = 625T(n/25) + 2n^2
= 25^2 * T( n/ ( 5^2 ) ) + 2 * n*n
= 625(25T(n/125)+n^2/625) + 2n^2
= 625*25*T(n/125) + 3n^2
= 25^3 * T( n/ ( 5^3 ) ) + 3 * n*n
= ....
= 25 ^ x * T( n / 5^x ) + x * n^2
T(n) = 25T(n/5)+n^2
T(0) = 25T(0) + n^2 ==> T(0) = 0
T(1) = 25T(0)+n^2 ==> T(1) = 1
x = lg 5 n
25 ^ x * T( n / 5^x ) + x * n^2
= n^2 * 1 + lg 5 n * n^2
= n^2*(lgn)

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