A. Mean Inequality
Solution:
Sort the array. And split the array to two parts. Each time, we take one number from each of the two parts at a time.
Code:
Java
C++
B. I Hate 1111
Solution:
Consider 1111, we found that $1111=11\times 101$. In same way, we only need consider if the number can be built by 11 and 111.
Let $x=a\cdot 11 + b\cdot 111$. when $b>11$, $x=a\cdot 11 + 11 \cdot 111 + (b-11)\cdot 111=(a+111)\cdot 11 + (b-11) \cdot 111$. So, if there is a solution, then we must can find the solution which $b<11$. So, just enumerate the b.
Code:
Java
C++
C. Potions
Tried to find some $n^2$ solution, so that can pass easy version only. But I cannot find that.
And for this problem, it is obviously, just calculating the prefix sum. When the sum is less than 0, we spit the worst potion.
D. Kill Anton
The conclusion is: in the answer, the four letters must be consecutive. So just enumerate the permutation of four letters and see which one can make largest cost.
E. Oolimry and Suffix Array
After I build the 4th sample data, then this problem is solved. The point is: find out the way to handle the case which string have duplicate character.
