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Golfed, I get you down to 140, assuming the inputs are ordered smallest to largest. Interestingly enough, if you have the defined-or patch, I get down to 122. If the inputs aren't ordered, add 13 characters to both solutions.
 ```Without //= patch: @x=@_;my%c;\$e=sub{my\$v=pop;exists\$c{\$v}?\$c{\$v}:\$c{\$v}=\$v<0?0:\$v==0||gr +ep&\$e(\$v-\$_),@x};\$t=\$s=0;{&\$e(++\$t)?\$t-\$s>=\$x[0]&&last:(\$s=\$t);redo}\$ +s With //= patch: @x=@_;my%c;\$e=sub{my\$v=pop;\$c{\$v}//=\$v<0?0:\$v==0||grep&\$e(\$v-\$_),@x};\$ +t=\$s=0;{&\$e(++\$t)?\$t-\$s>=\$x[0]&&last:(\$s=\$t);redo}\$s [download]```

Update: Actually, the inputs don't have to be ordered. It just means that the algorithm will take a little longer, but it will get the right results. Also, drop a character by reordering the assignment to \$c{\$v}. The //= patched version looks like:
 ```@x=@_;my%c;\$e=sub{my\$v=pop;\$c{\$v}//=\$v==0||\$v>0&&grep&\$e(\$v-\$_),@x};\$t +=\$s=0;{&\$e(++\$t)?\$t-\$s>=\$x[0]&&last:(\$s=\$t);redo}\$s [download]```

Update: Rewriting the redo-loop as a C-style for-loop drops to 111 characters for the //= patched version.
 ```@x=@_;my%c;\$e=sub{my\$v=pop;\$c{\$v}//=!\$v||\$v>0&&grep&\$e(\$v-\$_),@x};for( +\$t=\$s=0;\$t-\$s<\$x[0];&\$e(++\$t)or\$s=\$t){}\$s [download]```

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