[codility]Number-of-disc-intersections

// you can also use includes, for example:#include <algorithm>struct IntervalSeg{    long long left, right;    IntervalSeg(long long _left = 0, long long _right = 0):left(_left),right(_right){}    bool operator<(const IntervalSeg& orh) const    {        if(left == orh.left) return right < orh.right;        else return left < orh.left;    }};int getFirstBiggerIdx(const vector<IntervalSeg>& intervalSeg, long long target){    int left = 0;    int right = intervalSeg.size()-1;    while(left <= right)    {        int mid = left+(right-left)/2;        if(intervalSeg[mid].left > target) right = mid-1;        else left = mid+1;    }    return left;}int solution(const vector<int> &A) {    // write your code in C++98    //...convert original problem to counting overlap number of given interval segments    vector<IntervalSeg> intervalSeg(A.size());    for(int i = 0; i < A.size(); ++i)    {        intervalSeg[i].left = intervalSeg[i].right = i;        intervalSeg[i].left -= A[i];        intervalSeg[i].right += A[i];    }    //...first interval segments first    sort(intervalSeg.begin(), intervalSeg.end());    //...enumerate each interval segment and using binary search to calculate how much interval segments overlap it    long long result = 0;    for(int i = 0; i < intervalSeg.size(); ++i)    {        long long curRight = intervalSeg[i].right;        int firstBiggerIdx = getFirstBiggerIdx(intervalSeg, curRight);        result += firstBiggerIdx-i-1;    }    //...return result    long long MAX_LIMIT = 10000000;    if(result > MAX_LIMIT) return -1;    else return (int)result;}