SGU 109 Magic of David Copperfield II

走n步，先判断距离大于n的点，删掉

然后每次斜着删

#include<iostream>#include<vector>#include<algorithm>#include<cstdio>#include<queue>#include<stack>#include<string>#include<map>#include<set>#include<cmath>#include<cassert>#include<cstring>#include<iomanip>using namespace std;#ifdef _WIN32#define i64 __int64#define out64 "%I64d\n"#define in64 "%I64d"#else#define i64 long long#define out64 "%lld\n"#define in64 "%lld"#endif#define FOR(i,a,b)      for( int i = (a) ; i <= (b) ; i ++)#define FF(i,a)         for( int i = 0 ; i < (a) ; i ++)#define FFD(i,a)        for( int i = (a)-1 ; i >= 0 ; i --)#define S64(a)          scanf(in64,&a)#define SS(a)           scanf("%d",&a)#define LL(a)           ((a)<<1)#define RR(a)           (((a)<<1)+1)#define SZ(a)           ((int)a.size())#define PP(n,m,a)       puts("---");FF(i,n){FF(j,m)cout << a[i][j] << ' ';puts("");}#define pb              push_back#define CL(Q)           while(!Q.empty())Q.pop()#define MM(name,what)   memset(name,what,sizeof(name))#define read            freopen("in.txt","r",stdin)#define write           freopen("out.txt","w",stdout)const int inf = 0x3f3f3f3f;const i64 inf64 = 0x3f3f3f3f3f3f3f3fLL;const double oo = 10e9;const double eps = 10e-10;const double pi = acos(-1.0);const int maxn = 111;int n;int k;int step;inline int num(int x,int y){    return (x-1)*n+y;}int main(){    while(cin>>n)    {        if(n==2)        {            printf("3 4\n");            printf("5 2 3\n");        }        else        {            cout<<n;            FOR(i,1,n) FOR(j,1,n)            {                if(j+i-2 > n)                {                    cout<<" "<<num(i,j);                }            }            cout<<endl;            k = n + (n&1) + 1 ;            printf("%d",k);            FOR(i,1,n) FOR(j,1,n)            {                if(j+i-2 == n)                {                    cout<<" "<<num(i,j);                }            }            cout<<endl;            k+=2;            FOR(i,1,n-1)            {                printf("%d",k++);                k++;                FOR(j,1,n-i+1)                {                    cout<<" "<<num(n-i+1-j+1,j);                }                cout<<endl;            }        }    }    return 0;}