40 lines
1.1 KiB
C++
40 lines
1.1 KiB
C++
#include <bits/stdc++.h>
|
||
using namespace std;
|
||
const int N = 40010;
|
||
int n, m;
|
||
int q[N];
|
||
|
||
// 40分
|
||
bool check(int a, int b, int c, int d) {
|
||
if (a >= b || b >= c || c >= d) return 0;
|
||
if ((b - a) != 2 * (d - c)) return 0;
|
||
if (3 * (b - a) >= (c - b)) return 0;
|
||
return 1;
|
||
}
|
||
int g[N][4];
|
||
|
||
int main() {
|
||
#ifndef ONLINE_JUDGE
|
||
freopen("468.in", "r", stdin);
|
||
#endif
|
||
cin >> n >> m;
|
||
// 魔法值都是不超过n的正整数,似乎没啥用
|
||
// m个魔法物品
|
||
|
||
for (int i = 1; i <= m; i++) cin >> q[i]; // 读入每个魔法物品的魔法值
|
||
|
||
for (int a = 1; a <= m; a++)
|
||
for (int b = 1; b <= m; b++)
|
||
for (int c = 1; c <= m; c++)
|
||
for (int d = 1; d <= m; d++)
|
||
if (check(q[a], q[b], q[c], q[d]))
|
||
g[a][0]++, g[b][1]++, g[c][2]++, g[d][3]++;
|
||
|
||
// a这个枚举到的数字出现了一次，它是做为a位置出现的
|
||
// 找到一组合法的a,b,c,d
|
||
|
||
// 输出结果
|
||
for (int i = 1; i <= m; i++)
|
||
printf("%d %d %d %d\n", g[i][0], g[i][1], g[i][2], g[i][3]);
|
||
return 0;
|
||
} |