牛客网提高组Day2
T1 方差
第一眼看就知道要打暴力啊，然而并没有想到去化简式子。。。
可能因为昨晚没睡好，今天上午困死
导致暴力打了一个半小时，还不对。。。
#include <algorithm> #include <cstring> #include <cstdio> #include <cmath> using namespace std; typedef long long LL; const int M = 100005; int n, m; double sum; LL a[M], s[M], f[M]; double mul(double a, int b) { double res = 0.0; while (b) { if (b & 1) res += a; a += a; b >>= 1; } return 0; } int main() { scanf("%d", &n); m = n - 1; for (int i = 1; i <= n; i++) { scanf("%d", &a[i]); sum += a[i]; } for (int i = 1; i <= n; i++) { if (f[a[i]] != 0) { if(i != n) printf("%lld ", f[a[i]]); else return printf("%lld ", f[a[i]]), 0; continue; } double tmp = 1.0 * (sum - a[i]) / m; double ss; for (int j = 1; j <= n; j++) { if (i == j) continue; double t = abs(a[j] - tmp); t = 1.0 * t * t; ss += t; } f[a[i]] = mul(ss, m); if(i != n) printf("%lld ", f[a[i]]); else printf("%lld ", f[a[i]]); } return 0; }
正解：
题中公式可进行化简，转化为只需维护序列元素的和，平方和即可
#include<iostream> #include<cstdio> using namespace std; const int M = 100002; int n; long long s, s2, ans; long long a[M]; int main() { scanf("%d", &n); for (int i = 1; i <= n; i++) { scanf("%lld", &a[i]); s += a[i]; s2 += a[i] * a[i]; } for (int i = 1; i <= n; i++) { ans = (n - 1) * (s2 - a[i] * a[i]) - (s - a[i]) * (s - a[i]); printf("%lld ", ans); } return 0; }
T2 分糖果
暴力搜索 然而并没有分 qwq。。。
#include <algorithm> #include <iostream> #include <cstdio> using namespace std; const int M = 110; int n, ans; int f[M], a[M], vis[M]; void dfs(int step) { if(step == n + 1) ans++; else for(int j = 1; j <= f[step]; j++) { if(!vis[j] && step == n && (a[step - 1] != j) && a[1] != j) { a[step] = j; dfs(step + 1); } else if(!vis[j] && (a[step - 1] != j) && step != n) { a[step] = j; dfs(step + 1); } } } int main() { scanf("%d", &n); for(int i = 1; i <= n; i++) scanf("%d", &f[i]); dfs(1); printf("%d ", ans); return 0; }
正解：
暴力复杂度为O(n^2)，所以考虑优化：
线段树优化：O(nlogn) 80pts
单调队列优化：O(n) 100pts
#include<bits/stdc++.h> using namespace std; const int M = 1000005; const int P = 1e9 + 7; int n, B[M], A[M]; int dp[M], sum[2]; int q[2][M], val[2][M], l[2], r[2]; void insert(int f, int x, int s) { while (l[f] < r[f] && q[f][r[f] - 1] >= x) { r[f]--; s = (s + val[f][r[f]]) % P; sum[f] = (sum[f] - 1ll * val[f][r[f]] * q[f][r[f]] % P + P) % P; } val[f][r[f]] = s; q[f][r[f]++] = x; sum[f] = (sum[f] + 1ll * s * x % P) % P; s = 0; while (l[1 - f] < r[1 - f] && q[1 - f][r[1 - f] - 1] >= x) { r[1 - f]--; s = (s + val[1 - f][r[1 - f]]) % P; sum[1 - f] = (sum[1 - f] - 1ll * val[1 - f][r[1 - f]] * q[1 - f][r[1 - f]] % P + P) % P; } if (s) { val[1 - f][r[1 - f]] = s; q[1 - f][r[1 - f]++] = x; sum[1 - f] = (sum[1 - f] + 1ll * s * x % P) % P; } } int main() { scanf("%d", &n); int mi = 1, ans = 0; for (int i = 1; i <= n; i++) { scanf("%d", &B[i]); if (B[mi] > B[i])mi = i; } int len = 0; for (int i = mi; i <= n; i++) A[++len] = B[i]; for (int i = 1; i < mi; i++) A[++len] = B[i]; dp[0] = 1; l[0] = r[0] = l[1] = r[1] = 1; for (int i = 1; i <= n; i++) { int mi = A[i]; int f = i & 1; insert(i & 1, A[i], dp[i - 1]); dp[i] = (sum[f] - sum[1 - f] + P) % P; if (i > 1) ans = (dp[i] - ans + P) % P; } printf("%d ", ans); return 0; }
T3 集合划分
直接输出“-1”没有分 差评。。
正解：
#include<bits/stdc++.h> #define gc getchar() #define pc putchar using namespace std; typedef long long li; const int M = 300005; bool fg[M], vst[M], pt[20]; int n, m, k, mx, a[10010]; int h, t, ft, st[20]; int q[M], f[M], tj[M], lst[M]; int as[M]; li s1 = 19260817, s2 = 23333333; li s3 = 998244853, srd; li read() { li x = 0, y = 0, c = gc; while (!isdigit(c)) y = c, c = gc; while (isdigit(c)) x = (x << 1) + (x << 3) + (c ^ '0'), c = gc; return y == '-' ? -x : x; } void print(li q) { if (q < 0) { pc('-'); q = -q; } if (q >= 10) print(q / 10); pc(q % 10 + '0'); } li rd() { return srd = (srd * s1 + s2 + rand()) % s3; } int main() { srand(time(0)); rd(); int i, j, l; n = read(); m = read(); k = read(); mx = (1 << n) - 1; for (i = 1; i <= m; ++i) a[i] = read(), f[a[i]] = a[i]; if (m > k) return puts("-1"), 0; for (i = 1; i <= mx; ++i) tj[i] = tj[i >> 1] + (i & 1); for (i = 1; i <= mx; i <<= 1) for (j = 0; j <= mx; j += (i << 1)) for (l = j; l < j + i; ++l) f[l + i] |= f[l]; int q1 = 0, q2 = 0; for (i = 0; i <= mx; ++i) fg[i] = 1; for (i = 1; i <= n; ++i) { if (k & (1 << n - i)) ++q1; else ++q2; for (j = 0; j <= mx; ++j) if (n - tj[j] == q1 && tj[j] - tj[f[j]] < q2) fg[j] = 0; } if (!fg[mx]) { puts("-1"); return 0; } int nw, nxt; q[++t] = mx; vst[mx] = 1; while (h < t) { nw = q[++h]; for (i = 1; i <= n; ++i) if (nw & (1 << i - 1)) { nxt = nw ^ (1 << i - 1); if (!fg[nxt] || vst[nxt]) continue; vst[nxt] = 1; lst[nxt] = nw; q[++t] = nxt; } } if (!vst[0]) { puts("-1"); return 0; } for (nw = 0; nw != mx; nw = lst[nw]) st[++ft] = nw; st[++ft] = mx; for (i = 1; i <= n; ++i) { if (k & (1 << n - i)) { while (pt[ft - 1]) --ft; nw = st[ft] ^ st[ft - 1]; --ft; for (j = 1; j <= mx; ++j) if (!as[j] && (j & nw)) as[j] = 2; } else { l = 0; for (j = 1; j <= m; ++j) if (!as[a[j]]) l |= a[j]; for (j = 1; j < ft; ++j) if (!pt[j] && ((st[j] ^ st[j + 1]) & l) == 0) { nw = st[j] ^ st[j + 1]; pt[j] = 1; break; } for (j = 1; j <= mx; ++j) if (!as[j] && (j & nw)) as[j] = 1; } } for(i = 1; i <= mx; ++i) pc(as[i] - 1 + '0'); pc(' '); return 0; }