Careercup | Chapter 7

7.4 write methods to implement the multiply, subtract, and divide operations for integers. use only the Add operator.

Relatively simple. But it must be encapsulated.

7.5 Given two squares on a two-dimen1_plane, find a line that wocould cut these two squares in half. assume that the top and the bottom sides of the Square run parallel to the x-axis.

How to Write it concisely. Solution:

1. How to comprehensively consider various situations?

This type of question refers to the special case idea and then the algorithm.

7.6 given a two-dimenstmgraph with points on it, find a line which passes the most number of points.

See here. At that time, precision was not considered.

 1 int findMaxLine(vector<int> &points) { 2     int max = 0; 3     int dup = 0; 4     map<int, int> counts; 5     double epison = 0.0001; 6  7     for (int i = 0; i < points.size(); ++i) { 8         counts.clear(); 9         dup = 1;10         int m = 0;11         for (int j = i + 1; j < points.size(); ++j) {12             if (points[i].x == points[j].x && points[i].y == points[j].y) {13                 dup++;14             } else if (points[i].x == points[j].x) {15                 counts[0]++;16                 if (counts[0] > m) m = counts[0];17             } else {18                 double k = (points[i].y - points[j].y) * 1.0 / (points[i].x - points[j].x);19                 counts[(int)(k/epison)]++;20                 if (counts[int)(k/epison)] > m) m = counts[int)(k/epison)];21             }22         }23         if (m + dup > max) max = m + dup;24     }25     return max;26 }

7.7 design an algorithm to find the kth number such that the only prime factors are 3, 5, and 7.

 1 int findKthMagicNumber(int k) { 2     vector<queue<int> > queues(3); 3     queues[2].push(1); 4      5     for (int i = 0; i < k; ++i) { 6         int minIndex = 0, minNumber; 7         for (int j = 1; j < 3; ++j) { 8             if (!queues[j].empty() && queues[j].front() < queues[minIndex].front()) minIndex = j; 9         }10         minNumber = queues[minIndex].front();11         for (int j = minIndex; j < 3; ++j) {12             queues[j].push(minNumber * nums[j]);13         }14         queues[minIndex].pop();15     }16     return minNumber;17 }

 

Careercup | Chapter 7