51nod 1804 小C的多边形

a[i]=i

b[n-1]=1

b[]的其他值，可以直接算出来。

//#include<bits/stdc++.h>#include<stdio.h>#include<cstring>#include<cmath>#include<iostream>using namespace std;const int MAXN=1000100;int a[MAXN],b[MAXN];inline void prin_d(int x){    if (x > 9)    {        prin_d(x / 10);    }    putchar(x % 10 + '0');    return ;}int main(){int n,sum,i;while(~scanf("%d",&n)){if(n&1){printf("0\n");continue;}sum=3*n/2;n--;for(i=1;i<=n;i++)a[i]=i;b[n]=1;for(i=n-1;i>=1;i--)b[i]=sum-a[i]-b[i+1];prin_d(a[1]);for(i=2;i<=n;i++){printf(" ");prin_d(a[i]);}puts("");prin_d(b[1]);for(i=2;i<=n;i++){printf(" ");prin_d(b[i]);}puts("");}}