SGU 109 Magic of David Copperfield II

最新推荐文章于 2018-10-08 20:16:00 发布

最新推荐文章于 2018-10-08 20:16:00 发布 · 116 阅读

本文介绍了一个基于特定规则的算法实现，该算法通过判断点的距离并进行逐层删除操作来解决问题。代码使用C++编写，包括了标准输入输出、向量、算法等库，并定义了一系列辅助宏和函数来简化核心逻辑。

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;
}