A. Dense Array
Solution
If found the $\frac{max}{min}>2$ then just insert $2\cdot{min}$ in array.
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B. Balanced Remainders
Solution
Let find one of i which make $c_i>\frac{n}{3}$. Then move $c_i-\frac{n}{3}$ numbers to $c_{i+1}$. Then i++, and check if $c_i>\frac{n}{3}$ again.
The only small trap: this operation needs to be done 2 rounds. (which means do i++ 6 times.). Because in 1st round, the number of $c_i$ is not enough for $c_{i+1}$, most number is in $c_{i+2}$.
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C. Sum of Cubes
Solution
Brute force + binary search. Violence enumerates $a$ between 1 and 10000. Then binary search b for $x-a^3$.
There is another solution (I not tried this). Violence enumerates $a$. But get a inaccurate $b'$ directly by $\sqrt[3]{x-a^3}$. Then enumerate b from $b'-5$ to $b'+5$
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D. Permutation Transformation
Solution
Sort $a$. Then scan the $a$. For each $a[i]$ we scan now, it must be the new node of the tree. After that, we just sort $a$ by index again.
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E. Accidental Victory
Solution
Sort $a$. For each people $i$. let count the total tokens before $i$ (inclusive). Then just check if the total tokens can bigger or equal than people $i+1$. If so, can win. Otherwise fail.
But be careful. If some $i$ failed. Then all the people which $j<i$ failed.
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F. Equalize the Array
Solution
Count the number of occurrences of each number. And sort the count from largest to smallest. And then scan the code.
When scanning $i^{th}$ number, remove all the number after $i^{th}$ count. Because all these count smaller than $count[i]$. Only can occurrence 0 times.
Similarly, remove number before $i^{th}$ count until it occurrence only $count[i]$ times.
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G. Old Floppy Drive
Solution
binary search + implementation.
Firstly, let's focus on the case no need second round (back to 1). Then calculate prefix max value of $a$ to array $max$. So just binary search the first point which is larger or equal than $x$.
If one round is not enough. Then we check the sum of $a$. If $sum\leq{0}$, then no way, return -1.
If need more than one round and can have solution. Then find the first round $r$ that make $x-r\cdot{sum}\leq{max[n]}$. Then binary search $x-r\cdot{sum}-{max[n]}$ in $max$
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