PKU-CPC 2019 简要题解
2024-05-19 15:30:12
有的代码是现场写的可能比较难看，懒得改了……凑合看下吧。
找时间补补
A. Ball
分类讨论。
#include <bits/stdc++.h>#define LOG(FMT...) fprintf(stderr, FMT)using namespace std;typedef long long ll;struct Disjoint_Set_Union
{int father[200005];int size[200005];int edge[200005];int F[200005];void reset(int n){for(int i = 1; i <= n; i++)father[i] = i;fill(&size[1], &size[n]+1, 1);fill(&edge[1], &edge[n]+1, 0);fill(&F[1], &F[n]+1, 0);}int get_root(int i){if(father[i] == i)return i;elsereturn father[i] = get_root(father[i]);}
};int main()
{#ifndef ONLINE_JUDGEfreopen("test.in", "r", stdin);
#endifconst int P = 10000019;int t;scanf("%d", &t);static Disjoint_Set_Union dsu;static bool vis[200005];for(int _ = 1; _ <= t; _++) {int n, m;scanf("%d%d", &n, &m);dsu.reset(m);for(int _ = 1; _ <= n; _++) {int a, b;scanf("%d%d", &a, &b);if(a == b) {dsu.edge[dsu.get_root(a)]++;dsu.F[dsu.get_root(a)]++;}else if(dsu.get_root(a) == dsu.get_root(b)) {a = dsu.get_root(a);dsu.edge[a]++;}else {a = dsu.get_root(a);b = dsu.get_root(b);dsu.father[b] = a;dsu.size[a] += dsu.size[b];dsu.edge[a] += dsu.edge[b] + 1;dsu.F[a] += dsu.F[b];}}fill(&vis[0], &vis[m]+1, false);int ans = 1;for(int i = 1; i <= m; i++)if(! vis[dsu.get_root(i)]) {int j = dsu.get_root(i);vis[j] = true;if(dsu.edge[j] > dsu.size[j])ans = 0;else if(dsu.edge[j] == dsu.size[j] && ! dsu.F[j])ans = ans * 2 % P;else if(dsu.edge[j] == dsu.size[j] && dsu.F[j])ans = ans;elseans = (long long)ans * dsu.size[j] % P;}printf("%d\n", ans);}
}
B. Seal
C. Parade
类似均分纸牌，一个方向扫过去支付代价。
#include <bits/stdc++.h>#define LOG(FMT...) fprintf(stderr, FMT)using namespace std;typedef long long ll;const int N = 100010;int cnt[N][2];void solve() {memset(cnt, 0, sizeof(cnt));ll ans = 0;int n;scanf("%d", &n);for (int rep = 0; rep < n * 2; ++rep) {int x, y;scanf("%d%d", &x, &y);if (y < 1) {ans += 1 - y;y = 1;}if (y > 2) {ans += y - 2;y = 2;}if (x < 1) {ans += 1 - x;x = 1;}if (x > n) {ans += x - n;x = n;}++cnt[x][y - 1];}for (int i = 1; i <= n; ++i) {#define A cnt[i][0]
#define B cnt[i][1]--A;--B;if (A > 0 && B > 0) {ans += A + B;cnt[i + 1][0] += A;cnt[i + 1][1] += B;} else if (A <= 0 && B <= 0) {ans -= A + B;cnt[i + 1][0] += A;cnt[i + 1][1] += B;} else if (A <= 0 && B > 0) {if (-A > B) {ans += B;A += B;ans += abs(A);cnt[i + 1][0] += A;} else {ans += -A;B += A;ans += abs(B);cnt[i + 1][1] += B;}} else if (A > 0 && B <= 0) {if (A > -B) {ans += -B;A += B;ans += abs(A);cnt[i + 1][0] += A;} else {ans += A;B += A;ans += abs(B);cnt[i + 1][1] += B;}}}printf("%lld\n", ans);
}int main() {#ifndef ONLINE_JUDGEfreopen("test.in", "r", stdin);
#endifint t;scanf("%d",&t);while(t--)solve();
}
D. Circuit
FWT 板子。
#include <bits/stdc++.h>#define LOG(FMT...) fprintf(stderr, FMT)using namespace std;typedef long long ll;const int N = 15;int n, p, P;
int a[(1 << N) + 10], b[(1 << N) + 10];int norm(int x) { return x >= p ? x - p : x; }int Norm(int x) { return x >= P ? x - P : x; }void fwt(int* a) {for (int i = 0; i < n; ++i)for (int j = 0; j < 1 << n; ++j)if (!(j >> i & 1)) {int a0 = a[j], a1 = a[j | 1 << i];a[j] = Norm(a0 + a1);a[j | 1 << i] = Norm(a0 - a1 + P);}
}void deb(int* a) {for (int i = 0; i < 1 << n; ++i)LOG("%d ", a[i]);LOG("\n");
}void solve() {memset(a, 0, sizeof(a));memset(b, 0, sizeof(b));int x, y, z;scanf("%d%d%d%d%d", &n, &x, &y, &z, &p);P = p << n;while (x--) {int l, r, c;scanf("%d%d%d", &l, &r, &c);c = norm(c % p + p);a[l] = norm(a[l] + c);a[r + 1] = norm(a[r + 1] - c + p);}for (int i = 1; i < 1 << n; ++i) a[i] = norm(a[i] + a[i - 1]);while (y--) {int l, r, c;scanf("%d%d%d", &l, &r, &c);c = norm(c % p + p);b[l] = norm(b[l] + c);b[r + 1] = norm(b[r + 1] - c + p);}for (int i = 1; i < 1 << n; ++i) b[i] = norm(b[i] + b[i - 1]);fwt(a);fwt(b);for (int i = 0; i < 1 << n; ++i) a[i] = a[i] * (ll)b[i] % P;fwt(a);for (int i = 0; i < 1 << n; ++i) a[i] >>= n;for (int i = 1; i < 1 << n; ++i) a[i] = norm(a[i] + a[i - 1]);while (z--) {int l, r;scanf("%d%d", &l, &r);int ans = a[r];if (l)ans = norm(ans - a[l - 1] + p);printf("%d ", ans % p);}putchar('\n');
}int main()
{#ifndef ONLINE_JUDGEfreopen("test.in", "r", stdin);
#endifint t;scanf("%d", &t);while (t--)solve();
}
E. Coprime
容斥一下，vector 上二分。
#include <bits/stdc++.h>#define LOG(FMT...) fprintf(stderr, FMT)using namespace std;typedef long long ll;const int N = 100010;int n, m;
int a[N], mu[N];
vector<int> fac[N], ex[N];void pre(int n) {mu[1] = 1;for (int i = 1; i <= n; ++i) {for (int j = i + i; j <= n; j += i)mu[j] -= mu[i];}for (int i = 1; i <= n; ++i) {if (!mu[i]) continue;for (int j = i; j <= n; j += i)fac[j].push_back(i);}
}void solve() {fill(ex + 1, ex + 100001, vector<int>());scanf("%d%d", &n, &m);for (int i = 1; i <= n; ++i) {int x;scanf("%d", &x);for (int v : fac[x])ex[v].push_back(i);}int d = 0;while (m--) {int l, r, x;scanf("%d%d%d", &l, &r, &x);l ^= d; r ^= d; x ^= d;d = 0;for (int v : fac[x])d += mu[v] * (int)(upper_bound(ex[v].begin(), ex[v].end(), r) - lower_bound(ex[v].begin(), ex[v].end(), l));printf("%d\n", d);}
}int main()
{#ifndef ONLINE_JUDGEfreopen("test.in", "r", stdin);
#endifpre(100000);int t;scanf("%d", &t);while (t--) {solve();}
}
F. Graduation
暴力。
#include <bits/stdc++.h>#define LOG(FMT...) fprintf(stderr, FMT)using namespace std;typedef long long ll;struct Time
{int day;int l, r;
};int main()
{#ifndef ONLINE_JUDGEfreopen("test.in", "r", stdin);
#endifint t;scanf("%d", &t);for(int _ = 1; _ <= t; _++) {int already[6];for(int i = 1; i <= 5; i++)scanf("%d", &already[i]);int goal[6];for(int i = 1; i <= 5; i++) {scanf("%d", &goal[i]);goal[i] = max(goal[i] - already[i], 0);}int n;scanf("%d", &n);int id[11], type[11], d[11];vector<Time> time[11];for(int i = 1; i <= n; i++) {int k;scanf("%d%d%d%d", &id[i], &type[i], &d[i], &k);for(int _ = 1; _ <= k; _++) {int day, l, r;scanf("%d%d%d", &day, &l, &r);time[i].push_back({day, l, r});}}if(goal[1] + goal[2] + goal[3] + goal[4] + goal[5] > 25)printf("unable to graduate\n");else if(goal[1] == 0 && goal[2] == 0 && goal[3] == 0 && goal[4] == 0 && goal[5] == 0)printf("able to graduate\n");else {vector<int> ans(100);for(int sn = 2; sn < 1<<n+1; sn += 2) {int cnt[6] = {};for(int i = 1; i <= n; i++)if(sn & (1<<i))cnt[type[i]]++;if(cnt[3] > 1 || cnt[4] > 1 || cnt[5] > 1)continue;if(cnt[1] + cnt[2] + cnt[3] + cnt[4] + cnt[5] >= ans.size())continue;//----------------------------------//int score[6] = {};for(int i = 1; i <= n; i++)if(sn & (1<<i))score[type[i]] += d[i];if(score[1] + score[2] + score[3] + score[4] + score[5] > 25)continue;if(score[1] < goal[1] || score[2] < goal[2] || score[3] < goal[3] || score[4] < goal[4] || score[5] < goal[5])continue;//----------------------------------//bool vis[7][13] = {};for(int i = 1; i <= n; i++)if(sn & (1<<i))for(auto t : time[i])for(int i = t.l; i <= t.r; i++) {if(vis[t.day][i])goto TAG;elsevis[t.day][i] = true;}ans.clear();for(int i = 1; i <= n; i++)if(sn & (1<<i))ans.push_back(id[i]);TAG:;}if(ans.size() >= 98)printf("unable to graduate\n");else {sort(ans.begin(), ans.end());printf("able to graduate\n");for(auto i : ans)printf("%d ", i);printf("\n");}}}
}
G. Go and Oreo
H. Homomorphism
I. Chamber of Braziers
J. Matrix of Determinants
K. Winner Winner, Chicken Dinner!
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