P2119 [NOIP2016 普及组] 魔法阵
传送门
1
我们可以先写出\(O(m^4)\)的暴力
#include <bits/stdc++.h>
#define int long long
#define PII pair<int, int>
using namespace std;
const int inf = 0x3f3f3f3f;
const int MOD = 1e9 + 7, N = 4e4 + 5;
int n, m, ans[N][4]; struct Node
{ int x, id; friend bool operator< (Node x, Node y) { return x.x < y.x; }
}x[N]; signed main()
{ scanf("%lld%lld", &m, &n); for (int i = 1; i <= n; ++i) scanf("%lld", &x[i].x), x[i].id = i; sort(x + 1, x + 1 + n); for (int a = 1; a <= n; ++a) { for (int b = a + 1; b <= n; ++b) { if (x[a].x == x[b].x) continue; for (int c = n; c > b; --c) { if (4 * x[b].x >= 3 * x[a].x + x[c].x) break; if (x[b].x == x[c].x) continue; for (int d = c + 1; d <= n; ++d) { if (x[c].x == x[d].x) continue; if (x[b].x - x[a].x != 2 * (x[d].x - x[c].x)) continue; ++ans[x[a].id][0], ++ans[x[b].id][1], ++ans[x[c].id][2], ++ans[x[d].id][3]; } } } } for (int i = 1; i <= n; ++i) printf("%lld %lld %lld %lld\n", ans[i][0], ans[i][1], ans[i][2], ans[i][3]); return 0;
}
其实这份代码是优化过一点的
比如
for (int c = n; c > b; --c) if (4 * x[b].x >= 3 * x[a].x + x[c].x) break;
把\(c\)反着循环，\(c\)会越来越小，如果这条式子现在满足不了，以后也满足不了
分数\(65pts\)
2
接着我们可以看到\(n\)的数据范围比\(m\)小
就会想到计数数组
可写出\(O(n^4)\)的暴力
#include <bits/stdc++.h>
#define int long long
#define PII pair<int, int>
using namespace std;
const int inf = 0x3f3f3f3f;
const int MOD = 1e9 + 7, N = 4e4 + 5;
int n, m, ans[N][4], mp[N], x[N];signed main()
{// freopen("magic.in", "r", stdin);// freopen("magic.out", "w", stdout);scanf("%lld%lld", &m, &n);for (int i = 1; i <= n; ++i)scanf("%lld", &x[i]), ++mp[x[i]];for (int a = 1; a <= m; ++a){if (!mp[a]) continue;for (int b = a + 1; b <= m; ++b){if (!mp[b]) continue;for (int c = m; c > b; --c){if (!mp[c]) continue;if (4 * b >= 3 * a + c) break;for (int d = c + 1; d <= m; ++d){if (!mp[d]) continue;if (b - a != 2 * (d - c)) continue;int tmp = mp[a] * mp[b] * mp[c] * mp[d];ans[a][0] += tmp / mp[a], ans[b][1] += tmp / mp[b];ans[c][2] += tmp / mp[c], ans[d][3] += tmp / mp[d];}}}}for (int i = 1; i <= n; ++i)printf("%lld %lld %lld %lld\n", ans[x[i]][0], ans[x[i]][1], ans[x[i]][2], ans[x[i]][3]);return 0;
}
分数: \(75pts\)
3
因为题目中说了 \(X_b-X_a=2(X_d-X_c)\)
所以，我们可以只枚举\(X_a, X_b, X_c\)
#include <bits/stdc++.h>
#define int long long
#define PII pair<int, int>
using namespace std;
const int inf = 0x3f3f3f3f;
const int MOD = 1e9 + 7, N = 4e4 + 5;
int n, m, ans[N][4], mp[N], x[N];signed main()
{// freopen("magic.in", "r", stdin);// freopen("magic.out", "w", stdout);scanf("%lld%lld", &m, &n);for (int i = 1; i <= n; ++i)scanf("%lld", &x[i]), ++mp[x[i]];for (int a = 1; a <= m; ++a){if (!mp[a]) continue;for (int b = a + 1; b <= m; ++b){if (!mp[b]) continue;if ((b - a) % 2) continue; // 因为式子是(b - a + 2 * c) / 2，c * 2一定是偶数，那么 b - a 也得是偶数for (int c = m; c > b; --c){if (!mp[c]) continue;if (4 * b >= 3 * a + c) break;int d = (b - a + 2 * c) / 2;if (!mp[d]) continue;int tmp = mp[a] * mp[b] * mp[c] * mp[d];ans[a][0] += tmp / mp[a], ans[b][1] += tmp / mp[b];ans[c][2] += tmp / mp[c], ans[d][3] += tmp / mp[d];}}}for (int i = 1; i <= n; ++i)printf("%lld %lld %lld %lld\n", ans[x[i]][0], ans[x[i]][1], ans[x[i]][2], ans[x[i]][3]);return 0;
}
分数: \(85pts\)
4
我们可以知道
这样， 中间的\(>6x\)的数量可以用后缀和处理
\[\displaystyle
C[i]+=\sum_{j=1}^{j+8x<i}mp[j]\times mp[j+2t]\times mp[i+t] \\
C[i] += mp[i+t]*\sum^{j+8t<i}_{j=1}cnt[j]\times cnt[j+2t] \\
\]
#include <bits/stdc++.h>
#define int long long
#define PII pair<int, int>
using namespace std;
const int inf = 0x3f3f3f3f;
const int MOD = 1e9 + 7, N = 4e4 + 5;
int n, m, ans[N][4], mp[N], val[N]; signed main()
{ // freopen("magic.in", "r", stdin); // freopen("magic.out", "w", stdout); scanf("%lld%lld", &m, &n); for (int i = 1; i <= n; ++i) scanf("%lld", &val[i]), ++mp[val[i]]; for (int x = 1; 9 * x + 1 <= n; ++x) { int sum = 0; int x7 = x * 7, x8 = x * 8, x9 = x * 9, x2 = x * 2; for (int i = x9 + 2; i <= n; ++i) { sum += mp[i - x7 - 1] * mp[i - x9 - 1]; ans[i - x][2] += mp[i] * sum; ans[i][3] += mp[i - x] * sum; } sum = 0; for (int i = n - x9 - 1; i >= 1; --i) { sum += mp[i + x8 + 1] * mp[i + x9 + 1]; ans[i][0] += mp[i + x2] * sum; ans[i + x2][1] += mp[i] * sum; } } for (int i = 1; i <= n; ++i) printf("%lld %lld %lld %lld\n", ans[val[i]][0], ans[val[i]][1], ans[val[i]][2], ans[val[i]][3]); return 0;
}
