제출 #1261086

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1261086		SzymonKrzywda	질문 (CEOI14_question_grader)	C11	컴파일 에러	0 ms	0 KiB

encoder

using namespace std;

bool wygen = 0;
vector<int> liczby;

void gen() {
  wygen = 1;
  const int K = 1 << 12;
  for (int msk = 0; msk < K; msk++) {
    int ile = 0;
    for (int bit = 0; bit < 12; bit++) {
      if ((1 << bit) & msk) ile++;
    }
    if (ile == 6) liczby.push_back(msk);
  }
}


int encode (int n, int x, int y) {
  if (!wygen) gen();
  for (int i = 0; i < 12; i++) {
    if (((1 << i) & liczby[x]) && !((1 << i) & liczby[y])) return i + 1;
  }
}

decoder

using namespace std;

bool wygen = 0;
vector<int> liczby;

void gen() {
	wygen = 1;
	const int K = 1 << 12;
	for (int msk = 0; msk < K; msk++) {
		int ile = 0;
		for (int bit = 0; bit < 12; bit++) {
			if ((1 << bit) & msk) ile++;
		}
		if (ile == 6) liczby.push_back(msk);
	}
}


int decode (int n, int q, int h) {
	if (!wygen) gen();
	h--;
	if (q & (1 << h)) return 1;
	else return 0;
}

컴파일 시 표준 에러 (stderr) 메시지

# 1번째 컴파일 단계

encoder.c:1:1: error: unknown type name 'using'
    1 | using namespace std;
      | ^~~~~
encoder.c:1:17: error: expected '=', ',', ';', 'asm' or '__attribute__' before 'std'
    1 | using namespace std;
      |                 ^~~
encoder.c:3:1: error: unknown type name 'bool'
    3 | bool wygen = 0;
      | ^~~~
encoder.c:1:1: note: 'bool' is defined in header '<stdbool.h>'; did you forget to '#include <stdbool.h>'?
  +++ |+#include <stdbool.h>
    1 | using namespace std;
encoder.c:4:7: error: expected '=', ',', ';', 'asm' or '__attribute__' before '<' token
    4 | vector<int> liczby;
      |       ^
encoder.c: In function 'gen':
encoder.c:14:19: error: 'liczby' undeclared (first use in this function)
   14 |     if (ile == 6) liczby.push_back(msk);
      |                   ^~~~~~
encoder.c:14:19: note: each undeclared identifier is reported only once for each function it appears in
encoder.c: In function 'encode':
encoder.c:22:21: error: 'liczby' undeclared (first use in this function)
   22 |     if (((1 << i) & liczby[x]) && !((1 << i) & liczby[y])) return i + 1;
      |                     ^~~~~~