I've seen few people got AC in 0.015 seconds, but how? I used binary search and each iteration's left border started from the last found number, but not from the beginning of array and got 0.031 seconds.

binary search gives n*log^2(n) and its possible to solve this in 3*n if you walk the array with least number on each step and wait until all 3 pointed are same to increase count;
I have 0.031 too but i`m sure it is possible to optimize my code (e.g. i`ve placed too many checkers if the array is fully passed)

int main()
{
    int ma=0,mb=0,mc=0,an,bn,cn,a[4000],b[4000],c[4000],sum=0;

    scanf("%d",&an);
    for(int i=0;i<an;++i) scanf("%d",&a[i]);

    scanf("%d",&bn);
    for(int i=0;i<bn;++i) scanf("%d",&b[i]);

    scanf("%d",&cn);
    for(int i=0;i<cn;++i) scanf("%d",&c[i]);

    do
    {
        while ((a[ma]==b[mb])&&(b[mb]==c[mc]))
        {
            ++sum;
            ++ma;
            ++mb;
            ++mc;
            if((ma>=an)||(mb>=bn)||(mc>=cn)) break;
        }
        if((ma>=an)||(mb>=bn)||(mc>=cn)) break;

        if(a[ma]==min(min(b[mb],c[mc]),a[ma])) ++ma;
        else if(b[mb]==min(min(b[mb],c[mc]),a[ma])) ++mb;
        else if(c[mc]==min(min(b[mb],c[mc]),a[ma])) ++mc;

        if((ma>=an)||(mb>=bn)||(mc>=cn)) break;
    } while(1);

    printf("%d", sum);
    return 0;
}

You use Python. There is many useful functions))) My program took only 4 lines)))