int total(const vector<int> &A){    int ret = 0;    for (size_t i = 0; i < A.size(); ++i)        ret += A;    return ret;}int equi ( const vector<int> &A ) {    if (A.empty()) return -1;    if (A.size() == 1) return 0;    int right = total(A);    int left = 0;    for (size_t i = 0; i < A.size(); ++i)    {        right -= A;        if (left == right)            return i;        left += A;    }    return -1;}

Quote:
extreme_large_numbers
Sequence with extremly large numbers testing arithmetic overflow. 0.004 s. WRONG ANSWER
got 2, but it is not equilibrium point, sum[0..1]=4294967294, sum[3..3]=-2

Quote: Original post by nullsquared
I didn't think to test for overflow, that is such an obscure case [wow]... Python doesn't have that issue, does it?

Quote: Original post by nullsquared
I didn't think to test for overflow, that is such an obscure case [wow]... Python doesn't have that issue, does it?

Quote: Original post by benryves

Quote: Original post by nullsquared
I didn't think to test for overflow, that is such an obscure case [wow]... Python doesn't have that issue, does it?

I guess not, I also failed that test.

Quote: Original post by nullsquared

Quote: Original post by benryves

Quote: Original post by nullsquared
I didn't think to test for overflow, that is such an obscure case [wow]... Python doesn't have that issue, does it?

I guess not, I also failed that test.

What score did you get? I'm curious if the elapsed time has any effect on the score.

Quote: Original post by Sneftel
EDIT: I think I actually prefer crusadingknight's approach, since it doesn't involve any special-casing for the last iteration.

def equi(A):    total = sum(A)    accum = 0    for idx in range(len(A)):        if accum == total - accum - A[idx]:            return idx        accum += A[idx]    return -1

Quote: Original post by benryves
I don't think time matters as long as it doesn't time out.

#include <numeric>int equi ( const vector<int> &A ) {    if (A.empty()) return -1;    if (A.size() == 1) return 0;    int right = std::accumulate(A.begin(), A.end(), 0);    int left = 0;    for (size_t i = 0; i < A.size(); ++i)    {        right -= A;        if (left == right)            return i;        if (left + A < left && A > 0)            return -1;        left += A;    }    return -1;}

Quote: Original post by crusadingknight

def equi ( A ):  a = 0  b = sum(A)  for i in range(len(A)):    b -= A    if a == b:      return i    a += A  return -1