Thank you for registering 100,000 paiza.

What is this

There was a paiza 100,000 registration thank you special W campaign. It seems that it is popular to summarize, so both me and me. By the way, my uncle is only in C language.

algorithm

--Prepare a big array --Mark the loaded s_i * s_j --Count from p_i and output the number advanced to the mark

code

I wrote it down and didn't know which one I submitted (ぉ) Probably this one.

i,j,S[1<<20],F[4096];
main(X)
{
    for(;~scanf("%d",F+i++);)
        for(j=*S=*F+2;j<i;S[X<S-F?X:0]=1)
            X=F[i-1]*F[j++];

    for(i=1;++i-*S;printf("%d\n",j-X))
        for(X=j=F[i];!S[j];++j);
}

It's basically the same as fuyutsubaki.

A snake foot named commentary

• Read M, N, p \ _m, s \ _n into array F.
• M = F [0], N = F [1], p \ _m = F [2 .. F [0] +2], s \ _n = F [F [0] +2 .. F [1] It will be in +2]. (\ * S is subscripted with F [0] +2 == s_0)
• While reading, calculate X = s_i \ * s_j and set S [X] to 1 if it falls into the big array S. (To check if it fits, I calculate the size of the big array S with SF, but it seems that it is the reverse (FS), but it can not be helped because the gcc at hand drops that code. .)
• For p_m, count and output the number from the big array S [p_m] until a non-zero value appears.

Self-criticism

• At first, I thought about the code size as "the number of bytes in the executable file" and rewrote printf to putchar, and thought too much.
• I was too particular about 1000000.
• I thought too much about taking advantage of the side effects with the qsort function.
• Treating the array as char and trying to find the number up to the mark with the strlen function, I thought too much.
• If s_i and s_j are elementary, the Chinese Remainder Theorem can be used, right? I thought, and my interest turned to a side street.
• I wasn't a basic golfer, but I couldn't do it too much.

That's all from the field.