PAT-BASIC1041——考試座位號
阿新 • • 發佈:2018-12-17
題目描述:
知識點:字串
思路:開一個1001大小的陣列,試機座位號碼即陣列索引,陣列中儲存的值為考試座位號碼
本題是典型的空間換時間的思想。時間複雜度是O(1)。空間複雜度是O(1001)。
C++程式碼:
#include<iostream> #include<string> using namespace std; struct student { string studentNumber; int tryNumber; int testNumber; }; int main() { int totalNum; student tempStudent; string tempStudentNumber; int tempTryNumber; int tempTestNumber; cin >> totalNum; student students[1001]; for (int i = 0; i < totalNum; i++) { cin >> tempStudentNumber >> tempTryNumber >> tempTestNumber; tempStudent.studentNumber = tempStudentNumber; tempStudent.tryNumber = tempTryNumber; tempStudent.testNumber = tempTestNumber; students[tempTryNumber] = tempStudent; } int queryNum; int queryNumber; cin >> queryNum; for (int i = 0; i < queryNum; i++) { cin >> queryNumber; cout << students[queryNumber].studentNumber << " " << students[queryNumber].testNumber << endl; } }
C++解題報告: