Exercises on codility (14)

(1) Tieropes

Given n-segment rope--a positive integer array, and a positive integer k, can only connect adjacent two ropes at a time, connect the length of the rope to the length of the rope before, and the position is unchanged, asked so connected, up to how many root length of at least k rope?

Data range: n[1..10^5], array elements and K range [1..10^9].

Complexity required: Time O (N), Space O (1).

Analysis: Assuming that a rope is eventually thrown away, why not attach the rope to its adjacent rope? So I won't throw the rope ... So the linear sweep sum >= k is a ...

You can also with includes, for example://#include <algorithm>int solution (int K, vector<int> &a) {
   
    //Write your code in c++11    int r = 0;    for (int i = 0; i < a.size ();) {        int length = 0;        for (; (I < A.size ()) && (length < K); Length + = a[i++])        ;        if (length >= K) {            ++r;        }    }    return r;    }
   

(2) Maxnonoverlappingsegments

Given n segments, each segment is [A[i],b[i]] (closed interval), and the line segment has been sorted by the end endpoint to find out how many segments with no common points can be selected.

Data range N [0..30000], A, b arrays are integers, range [0..10^9].

Complexity of requirements: the time space is O (1).

Analysis: This is the event scheduling problem ... And the interval is sorted by the right endpoint, greedy one by one, the intersection will be thrown away.

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;int solution (vector<int> &a, vector<int> &b) {    //write your code in c++11    int last =-1, answer = 0;    for (int i = 0; i < a.size (), ++i) {        if (last < 0) | | (A[i] > B[last])) {Last            = i;            ++answer;        }    }    return answer;}

Exercises on codility (14)