SW ExpertAcademy/D2
1970. 쉬운 거스름돈
Programmer.
2018. 12. 1. 02:04
1970. 쉬운 거스름돈
풀이 방법
그리디 알고리즘에 따라 큰 화폐로 먼져 나눠 주고 그 다음 작은 화폐노 나눠주는것을 반복한다. 처음엔 막 썼지만 조금 더 깔끔하게 두 번쨰로 짜봤다.
<코드>
#include<iostream> using namespace std; int main(int argc, char** argv) { int test_case; int T; cin >> T; for (test_case = 1; test_case <= T; ++test_case) { unsigned int data{ 0 }; std::cin >> data; int count50000{ 0 }; int count10000{ 0 }; int count5000{ 0 }; int count1000{ 0 }; int count500{ 0 }; int count100{ 0 }; int count50{ 0 }; int count10{ 0 }; count50000 = data / 50000; if (count50000 > 0) data -= count50000 * 50000; count10000 = data / 10000; if (count10000 > 0) data -= count10000 * 10000; count5000 = data / 5000; if (count5000 > 0) data -= count5000 * 5000; count1000 = data / 1000; if (count1000 > 0) data -= count1000 * 1000; count500 = data / 500; if (count1000 > 0) data -= count500 * 500; count100 = data / 100; if (count100 > 0) data -= count100 * 100; count50 = data / 50; if (count50 > 0) data -= count50 * 50; count10 = data / 10; if (count10 > 0) data -= count10 * 10; std::cout << "#" << test_case << "\n" << count50000 << " " << count10000 << " " << count5000 << " " << count1000 << " " << count500 << " " << count100 << " " << count50 << " " << count10 << "\n"; } return 0;//정상종료시 반드시 0을 리턴해야합니다. }
두 번째
#include <iostream> #include <cstring> using namespace std; const int money[8]{ 50000,10000,5000,1000,500,100,50,10 }; int main(int argc, char** argv) { std::ios::sync_with_stdio(false); std::cout.tie(nullptr); std::cin.tie(nullptr); int test_case, T; cin >> T; for (test_case = 1; test_case <= T; ++test_case) { int result[8]; memset(result, 0, sizeof(result)); int input; std::cin >> input; for (int i = 0; i < 8; i++) { if (input == 0) break; if (money[i] <= input) { result[i] = input / money[i]; input = input % money[i]; } } std::cout << "#" << test_case << "\n"; for (int i = 0; i < 8; i++) std::cout << result[i] << " "; std::cout << "\n"; } return 0;//정상종료시 반드시 0을 리턴해야합니다. }