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-8
# define MOD 1000000007
# define INF 1000000000
# 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<<1,l,mid
# define rch p<<1|1,mid+1,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=0,f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
inline void Out(int a) {
    if(a<0) {putchar(‘-‘); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+‘0‘);
}
const int N=105;
//Code begin...

int main ()
{
    int n;
    scanf("%d",&n);
    if (n&1) {puts("0"); return 0;}
    int k=(n-1)/2;
    for (int i=n-1; i>=1; --i) Out(i), putchar(‘ ‘);
    putchar(‘\n‘);
    for (int i=k+1; i<2*k+1; ++i) Out(i), putchar(‘ ‘), Out(i-k), putchar(‘ ‘);
    Out(2*k+1); putchar(‘\n‘);
    return 0;
}

mycode