When matrix A is symmetric
$\sum_{i=1}^{N-1} \sum_{j=i+1}^N A_{ij} = \left(\sum_{ij} A_{ij} - \sum_i A_{ii}\right) / 2$
<ul>
<li>Specific examples <a href="/en/ABC177C">ABC177C</a>.<ul>
<li>$A_{ij} = A_iA_j$ when $\sum_{i=1}^{N-1} \sum_{j=i+1}^N A_iA_j = \left(\sum_{ij} A_iA_j - \sum_i A_i^2\right) / 2$<ul>
<li>Note that $\sum_{ij} A_iA_j = \left(\sum A_i\right)^2$</li>
</ul>
</li>
</ul>
</li>
</ul>
Especially when the diagonal component is 0
<ul>
<li>$\sum_{i=1}^{N-1} \sum_{j=i+1}^N A_{ij} = \left(\sum_{ij} A_{ij}\right) / 2$</li>
<li>Specific examples <a href="/en/ABC194">ABC194</a>C<ul>
<li>$A_{ij} = (A_i - A_j)^2$ when $\sum_{i=1}^{N-1} \sum_{j=i+1}^N A_{ij} = \left(\sum_{ij} (A_i - A_j)^2 \right) / 2$</li>
</ul>
</li>
<li><a href="/en/ABC147D">ABC147D</a><ul>
<li>$A_{ij} = A_i \oplus A_j$</li>
</ul>
</li>
</ul>
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