Question

还需要一个条件，即用户无法输入高于15的值。如果用户再次插入相同的六进制，它将重新启动＆＃34;循环＆＃34;

 int matrixArray[4][4], i, j, rowsum, columnsum, diagonalsum;


 printf("Skriv in 16 olika värden för din 4 x 4 kvadrat: "); // asks user to insert 16 numbers to see if its a magicsquare

 for (i = 0; i < 4; i++) {
        for (j = 0; j < 4; j++) {
            scanf_s("%x", &matrixArray[i][j]);  // inserts values into the array

            }
        }

Answer 1

我定义了一个＆＃34;看过＆＃34;的数组。数字，其中存储用户已输入的数字的0/1标志，如

int seen[16]; // initialize with 0

然后，每次scanf_s为您提供最新的输入（为简洁起见，我们称之为in），您可以检查seen[in] == 0。如果是，只需将号码存储在matrixArray中，请设置seen[in] = 1并继续。如果不是，则用户输入了两次号码。只需丢弃输入，给她一个很好的错误信息并继续。这可以在while循环中完成。

伪代码

ok = false
for i = 1..4 {
  for j = 1..4 {
    while (not ok) {
      in = scanf()
      if (not 1 <= in <= 16)
        error("input must be between 1 and 16; try again")
        continue

      if seen[in] == 1
        error("you already entered that number; try again")
        continue
      else
        matrixArray[i][j] = in
        seen[in] = 1
        ok = true
    }
  }
}

Answer 2

使用先前读取的值来查找重复项

#define ROW 4
#define COL 4
int total = ROW*COL;
printf("Skriv in %d olika värden för din %d x %d kvadrat: ", total, ROW, COL); 
for (int i = 0; i < total; i++) {
  bool duplicate;
  do {
    duplicate = false;
    unsigned value; 
    if (scanf_s("%x", &value) != 1) {
      printf("Numeric data not entered\n");
      exit(-1);
    }
    for (int k = 0; k < i; k++) {  // look through prior values
      int r2 = k/COL;
      int c2 = k%COL;
      if (value == matrixArray[r2][c2]) {
        duplicate = true;
        printf("Duplicate\n");
        break;
      }
    }
  } while (duplicate);  // reject recent entry and try again.
  int r = i/COL;
  int c = i%COL;
  matrixArray[r][c] = value;
}

防止用户多次输入相同的号码

2 个答案: