Question

It s difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.

Closed 11 years ago.

我的同事给出了以下文字值:1 2 3 4 . 15 ....... 4 2 1, 仅限、no function、no goto statements和, 不得使用任何有条件声明或私人操作者<>。

因此,我为解决这一问题而进行播种,但这不是一个准确的解决办法,因为15份不是两次印刷。

int main() { int i, j; for(i = 1, j = 0;j < 29;j++, i += int(j/15)*-2 + 1) cout<<i<<endl; }

<>Output:1 2 3 4 ... 15 14 13 ...... 2 1

任何替代解决办法?

Answer 1

您可以从1到30岁,然后使用以下事实(i/16)将“0”用于您的赞助部分,而“1”则用于您的赞助部分。

for (int i = 1; i < 31; i++)
{
    int number = (1-i/16) * i + (i/16) * (31 - i);
    printf("%d ", number);
}

Answer 2

for (int i=0; i<1; i++)
{
    std::cout << "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1"
}

Answer 3

如何做到这一点:

std::string first;
std::string second;

for ( int i = 1 ; i <= 15 ; i++ )
{
   std::ostringstream s;
   s << i;
   first += s.str();
   second = s.str() + second;
}

std::cout << first << second;

Answer 4

Alternative:

static int bla[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1};

for (int i = 0; i < 30; i++) 
{        
    printf("%d
", bla[i]);
}

情况良好,执行速度快于所有......

Answer 5

XOR bit #4 (i.e. j & 0x10) with bits 3:0. 你们需要找到一种办法,把这比喻为4个职位。

Answer 6

for (int i=1;i<31;++i)
{
  cout<<(((i<<27>>31|i)&(~i<<27>>31|~i))&15)<<" ";
}

Answer 7

#include <iostream>

int main()
{
    for(int i = 1; i < 31; i++) std::cout << ((i/16)-1)*-i+(i/16)*(i^0x1F) << " ";
    std::cout << std::endl;
}

Answer 8

const int N = 15;
for(int i = 1; i <= 2 * N; ++i)
    printf("%d ", i + (i > N) * (1 + 2 * (N - i)));

Answer 9

我看到许多复制<>/em>答案,但没有人利用作为。

std::string head = "1"; std::string tail = "1"; for (unsigned i = 2; i != 16; ++i) { std::string const elem = boost::lexical_cast<std::string>(i); head = head + " " + elem; tail = elem + " " + tail; } std::cout << head << " " << tail << " ";

在 ideone (minus lexical_cast):

A. 导言

它只是为了达到最高层(就你的电脑来说是足够的记忆)而工作。

Answer 10

for (int i = 1; i < 30; i++)
    printf("%d
", (-((i & 16) >> 4) + 1) * i + ((i & 16) >> 4) * (14 - (i & 15)));

Answer 11

int main()
{

    for(int i = 15, j = 30, k = 15; i <= 30; i++, j--, k -= 2)
    {
        cout << (j - i) * (k % 2) << endl << (j - i - 1) * (k % 2) << endl;
    }

    return 0;
}

这是我提出的。换言之,即:15 ->0 ->15。更多的思考食物。如果是负数,则使用 mo和 k。 I - J在中间举行会议。它绝不是完美的,还有其他一些更好的答案。

友情链接