Two interesting questions related to sequence

(4) The absolute value of the two numbers found in the array is minimized

Does it look like 2-sum? Not much to explain. Mainly is the absolute value big move can. It's a good way to sweep both ends!

Of course first sort, go out sort is O (n). Count on sort words degenerate to O (NLOGN)

This is also the problem of codility, before the time to tidy up.

Previous code:

You can also with includes, for example://#include <algorithm> #include <algorithm>int ab (int x) {    retur N (x >= 0)?
X: (-X);} int f (int x,int y) {    return ab (x + y);} int solution (vector<int> &a) {    //write your code in c++98    vector<int> a = A;    Sort (A.begin (), A.end ());    int answer = f (a[0],a[0]);     for (int i = 0, j = a.size ()-1; I <= J;) {        answer = min (answer, F (a[i], a[j]));        if (AB (a[i) > AB (A[j])) {            ++i;        }        else {            --j;        }            }    return answer;}

(5) A non-negative real array is given. has been ordered in accordance with non-descending order.

Set the array to C. The logarithm of the subscript length is N,0≤p < Q < N and C[p] * C[q]≥c[p] + c[q].

The question on codility is this definition of c[i] = A[i] + b[i]/10^6,a is the integer part [0..1000], and B is the numerator of the fractional part [0..999999] (the denominator of fractional portions is 1000000). Time Complexity of O (N)

The problem is actually analysis to be careful, again emphasizing that several numbers do not necessarily enumerate. We're just going to calculate a * b >= a + b because it's all positive. Again by symmetry we only consider the situation of 0<=a<=b.

Consider a larger number B possible scenario

(1) b = = 0 only A = = 0 to meet the conditions

(2) 0 < b < 2 no solution

(3) B >= 2 b/(b-1) <= a <= b

Note that assuming we are from small to large enumeration B, the condition (3) that b/(b-1) = 1/(1-1/b) is monotonically reduced, so for larger B. When we consider a, the previous legal A is also legal. This is the key to O (N), and we just need to record the last valid position of a.

On the code:

You can use includes, for example://#include <algorithm>//you can write to stdout for debugging purposes, e.g./ /cout << "This is a debug message" << endl;const int M = 1000000000;const int W = 1000000;long Long cmp (long   Long X1,long long y1, long long x2, long long y2) {//X1/y1-x2/y2 return x1 * y2-x2 * y1; }int solution (vector<int> &a, vector<int> &b) {//write your code in C++11/* Let A < = b A * b >= A + b b = = 0 A = = 0 0 < b < 2 no solution B >= 2 b/(b-1)    <= a <= b */int n = a.size (), NUM0 = 0, last = N, answer = 0;        for (int i = 0; i < n; ++i) {if ((a[i] = = 0) && (b[i] = = 0)) {//B = = 0 Answer + = num0++;            } else if (A[i] >= 2) {//b >= 2 if (last >= n) {last = i-1;            } int x = a[i] * W + b[i], y = x-w; for (; (last >= 0) && (CMP (a[last] * W + b[last],w, x, y) >= 0);            --last);        Answer + = i-1-last;        } if (answer >= m) {return m;            }//printf ("%d%d\n", I, answer);        } return answer; }

Two interesting questions related to sequence