51nod 1804 小C的多边形(构造)

首先可以算出无解的充分不必要条件，所有边的和为sum=3*((n-1)*n)/2，如果sum%n!=0显然无解。

也就是说n为奇数必然无解。现在考虑n为偶数的情况。

不妨假设n为偶数有解，现在考虑如何将这个解构造出来。

设此时n边形的为2*k+1,那么也就说，内边的每相邻两个边的和要为{k+2....3*k+2}

把边构造成这个样子即可。

k+1 1 k+2 2 k+3......2*k+1.

另外这个oj的读入效率真是感人肺腑啊。。。

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <bitset>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-
# define MOD
# define INF
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<,l,mid
# define rch p<<|,mid+,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
inline int Scan() {
    int x=,f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x*f;
}
inline void Out(int a) {
    if(a<) {putchar('-'); a=-a;}
    if(a>=) Out(a/);
    putchar(a%+'');
}
const int N=;
//Code begin...

int main ()
{
    int n;
    scanf("%d",&n);
    if (n&) {puts(""); return ;}
    int k=(n-)/;
    for (int i=n-; i>=; --i) Out(i), putchar(' ');
    putchar('\n');
    for (int i=k+; i<*k+; ++i) Out(i), putchar(' '), Out(i-k), putchar(' ');
    Out(*k+); putchar('\n');
    return ;
}

mycode