Half of the queue

When matrix A is symmetric $\sum_{i = 1}^{N - 1} \sum_{j = i + 1}^{N} A_{ij} = (\sum_{ij} A_{ij} - \sum_{i} A_{ii}) /2$

Specific examples ABC177C.
- $A_{ij} = A_{i} A_{j}$ when $\sum_{i = 1}^{N - 1} \sum_{j = i + 1}^{N} A_{i} A_{j} = (\sum_{ij} A_{i} A_{j} - \sum_{i} A_{i}^{2}) /2$
  - Note that $\sum_{ij} A_{i} A_{j} = (\sum A_{i})^{2}$

Especially when the diagonal component is 0

$\sum_{i = 1}^{N - 1} \sum_{j = i + 1}^{N} A_{ij} = (\sum_{ij} A_{ij}) /2$
Specific examples ABC194C
- $A_{ij} = (A_{i} - A_{j})^{2}$ when $\sum_{i = 1}^{N - 1} \sum_{j = i + 1}^{N} A_{ij} = (\sum_{ij} (A_{i} - A_{j})^{2}) /2$
ABC147D
- $A_{ij} = A_{i} \oplus A_{j}$

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Half of the queue

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