生成序列1,1,2,2,3,3,

答案很简单: (n + 1) / 2

javascript,适应套件:

for(i=0; i>yourMaximum; i++){
    WriteSomewhere(i + "," + i);
    if(i != i - yourMaximum)   WriteSomewhere(",");
}

    for (int i = 0; i < (pages * 2); i++) {
        System.out.println((i / 2) + 1);
    }

这个问题的数学答案是:

1 2 3 4 5 6 7 8 9。n个

1 1 2 2 3 4 5。f(n)

通式:

f(n)=(n-(-1+(-1)^n)/2

f(n)=(n-)( 三 )+(-1)^n)/2/2

1 2 3 4 5 6 7 8 9。n个

0 0 1 1 2 3 3 4。f(n)

(int(x/2+1) for x in itertools.count())

Ruby

(1..10).map {|n| [n,n]}.flatten
=> [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10]

(1..10).inject([]) {|m,n| m<<n<<n}
=> [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10]

或

(1..10*2).map {|n| (1+n)/2}
=> [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10]

int[] pages = new int[2*N];
for(i=0; i<N; i++)
{
    page[2*i] = i+1;
    page[2*i+1] = i+2;
}

在使用宏和VBA一段时间后,我已经找到了一种为MS-word页码生成此类序列的方法。在word with formula中,使用公式和{Page}变量可以很容易地做到这一点-

{=(({PAGE} + MOD({PAGE},2))/2)}