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zsh-89/209(Div.2)Permutation.cpp

Created November 3, 2013 15:09
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209(Div.2)Permutation.cpp
int main()
{
int n, k;
scanf("%d %d", &n, &k);
n *= 2;
for (int i = 1; i <= n; i+=2) {
if (k) {
--k;
printf("%d %d ", i, i+1);
} else
printf("%d %d ", i+1, i);
}
return 0;
}
@zsh-89
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zsh-89 commented Nov 3, 2013

有个更贪心的办法：
首先，如果要凑0，只要保证全部 ai > ai+1 或全部 ai < ai+1即可
如果要凑2，
2=4-2=(3+1)-(3-1)=|4-1| + |2-3| - |4-1+2-3|
4=8-4=(6+2)-(6-2)= ....
....
但是数字会越界，而且凑满接下来的0编程不方便。

这道题就是很考观察了，很有趣！

@eclipselu
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:lgtm:

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