【9707】循环比赛

Time Limit: 1 second
Memory Limit: 128 MB

【问题描述】

设有n个选手进行循环比赛，其中N=2^M,要求每名选手要与其他N-1名选手都赛一次，每名选手每天比赛一次，循环比赛共进行N-1天，
要求每天没有选手轮空。
【输入格式】

M

【输出格式】

表格形式的比赛安排表(每个数场宽为3）

Sample Input

2

Sample Output

1 2 3 4
2 1 4 3
3 4 1 2
4 3 2 1

【题目链接】:http://noi.qz5z.com/viewtask.asp?id=9707

【题解】

最小4格,左上和右下，右上和左下对称；
按照这个规则赋值整个表格就好;

【完整代码】

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <set>
#include <map>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
#include <stack>
#include <string>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

void rel(LL &r)
{
    r = 0;
    char t = getchar();
    while (!isdigit(t) && t!='-') t = getchar();
    LL sign = 1;
    if (t == '-')sign = -1;
    while (!isdigit(t)) t = getchar();
    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
    r = r*sign;
}

void rei(int &r)
{
    r = 0;
    char t = getchar();
    while (!isdigit(t)&&t!='-') t = getchar();
    int sign = 1;
    if (t == '-')sign = -1;
    while (!isdigit(t)) t = getchar();
    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
    r = r*sign;
}

const int MAXN = 2e3;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);

int a[MAXN][MAXN];
int m;

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    scanf("%d",&m);
    int half = 1;
    a[1][1] = 1;
    rep1(k,1,m)
    {
        rep1(i,1,half)
            rep1(j,1,half)
                a[i][j+half] = a[i][j]+half;
        rep1(i,1,half)
            rep1(j,1,half)
                {
                    a[i+half][j+half] = a[i][j];
                    a[i+half][j] = a[i][j+half];
                }
        half <<=1;
    }
    rep1(i,1,half)
        rep1(j,1,half)
            {
                printf("%3d",a[i][j]);
                if (j==half)
                    puts("");
            }
    return 0;
}