1- in -$k$ SAT to $k$-SAT reduction
Answer converted from comment.

Definition: 1-in-$k$ SAT is a $k$-SAT problem with the tighter condition that - given a truth assignment to the variables - each clause must contain exactly one true literal (and thus $k$-1 false literals).

See for example 1-in-three SAT on Wikipedia

The reduction from 1-in-$k$ SAT to $k$ SAT can be done easily.


following the definition, convert each $(X_1 \vee X_2 \vee \cdots \vee X_k)$ clause of the 1-in-$k$ SAT problem into $(X_1 \wedge -X_2 \wedge \cdots \wedge -X_k) \vee (-X_1 \wedge X_2 \wedge \cdots \wedge -X_k) \vee $ $\cdots \vee (-X_1 \wedge -X_2 \wedge \cdots \wedge X_k)$; 
convert to CNF distributing ORs over ANDs;
convert the resulting CNF formula to $k$ SAT


For example $(x_1 \vee x_2 \vee -x_3)$ becomes:


$(x_1 \wedge -x_2 \wedge x_3) \vee (-x_1 \wedge x_2 \wedge x_3) \vee
    (-x_1 \wedge -x_2 \wedge -x_3)$;
$(x_1 \vee x_2 \vee -x_3) \wedge (-x_1 \vee -x_2) \wedge (-x_1 \vee x_3) \wedge (-x_2 \vee x_3)$
$(x_1 \vee x_2 \vee -x_3) \wedge (-x_1 \vee -x_2 \vee x4) \wedge (-x_1 \vee -x_2 \vee -x4)$ $\wedge (-x_1 \vee x_3 \vee x4) \wedge (-x_1 \vee x_3 \vee -x4)$ $\wedge (-x_2 \vee x_3 \vee x4) \wedge (-x_2 \vee x_3 \vee -x4)$


Note: at step 3, if you allow repeated variables in a clause, then there is non need for the extra variable and the final $k$ SAT formula of the example can be:
$(x_1 \vee x_2 \vee -x_3) \wedge (-x_1 \vee -x_2 \vee -x2)$ $\wedge (-x_1 \vee x_3 \vee x3)$ $\wedge (-x_2 \vee x_3 \vee x3)$