SGU 109
题意:一个n*n的矩形,起点在1,1然后每次给你一个操作,走ki步,然后你可以删除任意一个点这次步走不到的,删了就不能再走了,然后问构造这种操作,使得最后删除n*n-1个点
剩下一个点,这个人最终的目的就在那,还要求每次走的步数要递增,n<=ki<300
收获:奇妙的构造,每次走奇数点,就会走到和自己奇偶不一样的点,(1,1)为偶点,(2,3)为奇点,然后第一次先删除n步走不到的点,那么接下来的点到起点的距离就是0-n,那么我们就可以
删n次,然后每一次删距离为n-i的点,最终他就会被逼到(1,1)了,太秀了
#include<bits/stdc++.h> #define de(x) cout<<#x<<"="<<x<<endl; #define dd(x) cout<<#x<<"="<<x<<" "; #define rep(i,a,b) for(int i=a;i<(b);++i) #define repd(i,a,b) for(int i=a;i>=(b);--i) #define repp(i,a,b,t) for(int i=a;i<(b);i+=t) #define ll long long #define mt(a,b) memset(a,b,sizeof(a)) #define fi first #define se second #define inf 0x3f3f3f3f #define INF 0x3f3f3f3f3f3f3f3f #define pii pair<int,int> #define pdd pair<double,double> #define pdi pair<double,int> #define mp(u,v) make_pair(u,v) #define sz(a) (int)a.size() #define ull unsigned long long #define ll long long #define pb push_back #define PI acos(-1.0) #define qc std::ios::sync_with_stdio(false) #define db double #define all(a) a.begin(),a.end() const int mod = 1e9+7; const int maxn = 1e2+5; const double eps = 1e-6; using namespace std; bool eq(const db &a, const db &b) { return fabs(a - b) < eps; } bool ls(const db &a, const db &b) { return a + eps < b; } bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); } ll gcd(ll a,ll b) { return a==0?b:gcd(b%a,a); }; ll lcm(ll a,ll b) { return a/gcd(a,b)*b; } ll kpow(ll a,ll b) {ll res=1;a%=mod; if(b<0) return 1; for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;} ll read(){ ll 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; } //inv[1]=1; //for(int i=2;i<=n;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod; int n,nn; int d[maxn][maxn]; void init(){ rep(i,1,n+1){ d[i][1] = i - 1; rep(j,2,n+1){ d[i][j] = d[i][j-1] + 1; } } } int main(){ scanf("%d",&n); nn = n; init(); if(n==2) return printf("3 4\n5 2 3\n"),0; printf("%d",n); rep(i,1,n+1) rep(j,1,n+1) if(d[i][j]>n) printf(" %d",(i-1)*n+j),d[i][j]=-1; puts(""); n = (n % 2?n + 2:n + 1); int dis = nn; rep(i,0,nn){ printf("%d",n); rep(j,1,nn+1) rep(k,1,nn+1) if(d[j][k]==dis) printf(" %d",(j-1)*nn+k); puts("");dis--;n += 2; } return 0; }