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;
}