5 7 1 2 1 2 1
16 Takahashi can cut the loaf for the children as follows. * Choose the loaf of length $7$ and $x=3$, cutting it into loaves of length $3$ and $4$ for a cost of $7$. * Choose the loaf of length $3$ and $x=1$, cutting it into loaves of length $1$ and $2$ for a cost of $3$. Give the former to the $1$\-st child. * Choose the loaf of length $2$ and $x=1$, cutting it into two loaves of length $1$ for a cost of $2$. Give them to the $3$\-rd and $5$\-th children. * Choose the loaf of length $4$ and $x=2$, cutting it into two loaves of length $2$ for a cost of $4$. Give them to the $2$\-nd and $4$\-th children. This incurs a cost of $7+3+2+4=16$, which is the minimum possible. There will be no leftover loaves.
3 1000000000000000 1000000000 1000000000 1000000000
1000005000000000 Note that each child $i$ must receive a loaf of length exactly $A_i$.
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