T1:
Seeking to meet $ (a + b) <= n $ and $ n | ab $ logarithmic number.
The answer is represented by the formula:
$\begin{array}{rl} ans &=& \sum \limits_{i=1}^n \sum \limits_{j=1}^n [i+j|ij][i+j<=n] \\ &=& \sum \limits_{i=1}^{n-1} \sum \limits_{j=1}^{n-i} [i+j|ij] \end{array}$
Provided $ g = gcd (a, b) $, when the $ \ frac {a + b} {g} | when g $, satisfy the conditions.
Proof:
If $ a + b | ab $, then $ \ frac {a + b} {gcd (a, b)} | \ frac {ab} {gcd (a, b)} $.
设$a'=\frac{a}{gcd(a,b)}，b'=\frac{b}{gcd(a,b)}$。
Provisions $ a '+ b' | a'b'gcd (a, b) $.
Because $ a '$ and $ b' $ prime, $ a '+ b' | gcd (a, b) $.
Logarithm to meet the conditions as follows:
$ans=\sum \limits_{i=1}^{n-1} \sum \limits_{j=1}^{n-i} [i+j|gcd(i,j)]$
Enumeration gcd:
$\begin{array}{rl} ans &=& \sum \limits_{g=1}^{\sqrt{n}} \sum \limits_{i=1}^{\lfloor \frac{n}{g} \rfloor} \sum \limits_{j=1}^{\lfloor \frac{n}{g} \rfloor -i}[gcd(i,j)==1][i+j|g] \\ &=& \sum \limits_{g=1}^{\sqrt{n}} \sum \limits_{i=2}^{\lfloor \frac{n}{g} \rfloor} \sum \limits_{j=1}^{i-1} [gcd(j,i-j)==1][i|g] \end{array}$
From the Decreases Technique:
$ Gcd (x, y) == gcd (and x) $
and so:
$\begin{array}{rl} ans &=& \sum \limits_{g=1}^{\sqrt{n}} \sum \limits_{i=2}^{\lfloor \frac{n}{g} \rfloor} [i|g] \sum \limits_{j=1}^{i-1} [gcd(i,j)==1] \\ &=& \sum \limits_{g=1}^{\sqrt{n}} \sum \limits_{i=1}^{\lfloor \frac{n}{g} \rfloor} [i|g] \varphi(i) \\ &=& \sum \limits_{g=1}^\sqrt{n} \sum \limits_{d|g} \varphi(d) \\ &=& \sum \limits_{d=1}^{\sqrt{n}} \lfloor \frac{n}{d^2}\rfloor \varphi(d )\end{array}$
Linear sieve out the $ \ varphi $ can.
Time complexity $ O (\ sqrt {n}) $.
T2:
Rise longest subsequence problem.
With the number of array shape optimized to achieve the complexity of $ nlogn $.
Take several statistical programs, the number of array-like structures are installed, continue to merge, find the number of programs under the optimal conditions of the current point, and then transferred.
Time complexity $ O (nlogn) $.
T3:
Fibonacci numbers, be sure to use Fibonacci number.
Have three arrays, $ dp [i] $ denotes the Fibonacci, $ f [i] $ I $ denotes the white point of the layer $, $ g [i] $ represents black points.
Three arrays can be obtained through a recursive, reprocessing and a prefix.
Points of discussion:
If lca in which a white point, the processing directly subtree and the white points, each layer into the white point.
lca nodes are black, enumeration path length, depth and a pitch lca, other points may be calculated in other cases.
You can then learn the height range of black nodes, multiplied by black dots can be.
Time complexity $ O (n ^ 2) $.