好题，用并查集代替普通 set 上二分的操作。
#include <bits/stdc++.h>
using namespace std;
typedef int ll;
const int N = 1e6 + 9;
struct Ranker : vector<ll>
{
Ranker() { reserve(N << 1); }
void init() { sort(begin(), end()), resize(unique(begin(), end()) - begin()); }
int ask(ll x)
{
return lower_bound(begin(), end(), x) - begin();
}
} rk;
struct UnionfindSet : vector<int>
{
UnionfindSet(int n) : vector<int>(n)
{
for (int i = 0; i < n; ++i)
at(i) = i;
}
void merge(int u, int w)
{
if (w = ask(w), u = ask(u), w != u)
at(w) = u;
}
int ask(int u) { return at(u) != u ? at(u) = ask(at(u)) : u; }
};
int n, q, z[N], x[N];
int main()
{
scanf("%d%d", &n, &q);
for (int i = 0; i < q; ++i)
{
scanf("%d%d", &z[i], &x[i]);
rk.push_back(x[i]);
if (x[i] + 1 <= n)
rk.push_back(x[i] + 1);
}
rk.init();
UnionfindSet ufs(rk.size());
for (int i = 0; i < q; ++i)
{
int t = rk.ask(x[i]);
if (z[i] == 1)
ufs.merge(t + 1, t);
else
printf("%d\n", rk[ufs.ask(t)]);
}
}
建议出题人重新学习一下英语，我希望出题人结合《Buy Watermelon》这题回答一下这个问题。
#include <bits/stdc++.h>
using namespace std;
int main()
{
int w;
scanf("%d", &w);
puts(w <= 2 || (w & 1) ? "NO" : "YES");
}
队友做的，似乎 kmp 搞一下就行了。
#include <cstdio>
#include <string.h>
char s1[500001], s2[500001];
int p[500001], ans = 0, q;
int main(void)
{
scanf("%s", s1 + 1);
scanf("%d", &q);
while (q--)
{
scanf("%s", s2 + 1);
ans = 0;
for (int i = 1; s2[i] != '\0' && s1[i] != '\0'; i++)
p[i] = 0;
p[1] = 0;
ans = 0;
int len1 = strlen(s1 + 1), len2 = strlen(s2 + 1), j = 0;
if (len1 > len2)
{
for (int i = 1; i < len2; i++)
{
while (j && s2[i + 1] != s2[j + 1])
j = p[j];
if (s2[i + 1] == s2[j + 1])
j++;
p[i + 1] = j;
}
j = 0;
for (int i = 0; i < len1; i++)
{
while (j && s1[i + 1] != s2[j + 1])
j = p[j];
if (s1[i + 1] == s2[j + 1])
j++;
if (j == len2)
{
ans = 1;
break;
j = 0;
}
if (ans)
break;
}
if (ans)
{
puts("my child!");
}
else
puts("oh, child!");
}
else if (len1 < len2)
{
p[1] = ans = 0;
for (int i = 1; i < len1; i++)
{
while (j && s1[i + 1] != s1[j + 1])
j = p[j];
if (s1[i + 1] == s1[j + 1])
j++;
p[i + 1] = j;
}
j = 0;
for (int i = 0; i < len2; i++)
{
while (j && s2[i + 1] != s1[j + 1])
j = p[j];
if (s2[i + 1] == s1[j + 1])
j++;
if (j == len1)
{
ans = 1;
j = 0;
break;
}
if (ans)
break;
}
if (ans)
{
puts("my teacher!");
}
else
puts("senior!");
}
else
{
int ct = 0;
for (int i = 1; i <= len2; i++)
{
if (s1[i] != s2[i])
{
ct = 1;
break;
}
}
if (ct == 0)
puts("jntm!");
else
puts("friend!");
}
}
}
CXK 打篮球了…按相反顺序插进去，树状数组维护一下前缀最大值即可。
#include <bits/stdc++.h>
using namespace std;
typedef int ll;
const int N = 5e5 + 9;
struct Ranker : vector<ll>
{
Ranker() { reserve(N << 1); }
void init() { sort(begin(), end()), resize(unique(begin(), end()) - begin()); }
int ask(ll x)
{
return lower_bound(begin(), end(), x) - begin();
}
} rk;
struct Fenwick : vector<int>
{
Fenwick(int N) : vector<int>(N + 9, -1) {}
void upd(int x, int val)
{
for (; x < size(); x += x & -x)
at(x) = max(at(x), val);
}
int ask(int x)
{
int r = -1;
for (; x; x -= x & -x)
r = max(r, at(x));
return r;
}
};
int n, m, w[N], ans[N];
int main()
{
scanf("%d%d", &n, &m);
for (int i = 0; i < n; ++i)
{
scanf("%d", &w[i]);
rk.push_back(w[i]);
rk.push_back(w[i] + m);
}
rk.init();
Fenwick t(rk.size() + 9);
for (int i = n - 1; ~i; --i)
{
int pos = rk.size() - rk.ask(w[i]), pos1 = rk.size() - rk.ask(w[i] + m);
t.upd(pos, i);
ans[i] = t.ask(pos1);
if (ans[i] >= 0)
ans[i] -= i + 1;
}
for (int i = 0; i < n; ++i)
printf("%d%c", ans[i], i < n - 1 ? ' ' : '\n');
}
队友用马拉车暴力搞掉了，集训队里也有用回文自动机做的？
#include <iostream>
#define maxl 300005
#define ll long long
using namespace std;
char s[maxl], s2[maxl << 1];
int R[maxl << 1], Rnow = -1, C, p2, cl, pl;
int oc[maxl][30];
int l = 0;
void init()
{
scanf("%s", s);
s2[l++] = '#';
for (int i = 0; s[i] != '\0'; i++)
s2[l++] = s[i], s2[l++] = '#';
for (int i = 0; s[i] != '\0'; i++)
for (int j = 0; j < 26; j++)
oc[i][j] = -1;
for (int i = 0; s[i] != '\0'; i++)
{
for (int j = 0; j < 26; j++)
{
if (s[i] - 'a' == j)
oc[i][j] = i;
else if (i != 0)
oc[i][j] = oc[i - 1][j];
}
}
}
void manacher()
{
for (int i = 0; i < l; i++)
{
if (i > Rnow)
{
for (int j = 0; i - j >= 0 && i + j < l && s2[i - j] == s2[i + j]; j++)
R[i] = j + 1;
Rnow = R[i] + i - 1;
C = i;
}
else
{
p2 = 2 * C - i;
pl = p2 - R[p2] + 1;
cl = 2 * C - Rnow;
if (pl > cl)
{
R[i] = R[p2];
}
if (pl == cl)
{
int tem = 2 * i - Rnow;
while (s2[Rnow + 1] == s2[tem - 1] && tem > 0 && Rnow < l - 1)
{
Rnow++, tem--;
}
R[i] = Rnow - i + 1;
C = i;
}
if (pl < cl)
{
R[i] = Rnow - i + 1;
}
}
//cout<<i<<" "<<C<<" "<<Rnow<<" "<<R[i]<<endl;
}
}
int main(void)
{
init();
manacher();
//for(int i=0;i<l;i++)cout<<s2[i];
ll ans = 0;
for (int i = 0; i < l; i++)
{
if (s2[i] != '#')
{
int l = i / 2 - R[i] / 2 + 1, r = i / 2 + R[i] / 2 - 1;
for (int j = 0; j < 26; j++)
{
int loc = oc[i / 2][j];
if (loc >= l)
ans += loc - l + 1;
//cout<<i<<" "<<j<<" "<<l<<" "<<loc<<endl;
}
}
else
{
if (i == 0 || i == l - 1)
continue;
int l = i / 2 - R[i] / 2, r = i / 2 - 2 + R[i] / 2;
for (int j = 0; j < 26; j++)
{
int loc = oc[i / 2 - 1][j];
if (loc >= l)
ans += loc - l + 1;
}
}
}
cout << ans;
}
注意这里对称是可以包含自己的，要特殊去处理一下。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1023;
int n;
map<pair<int, int>, int> mp;
set<pair<int, int>> se;
int main()
{
scanf("%d", &n);
for (int i = 0, x, y; i < n; ++i)
{
scanf("%d%d", &x, &y);
se.emplace(x, y);
}
int mx = 0;
vector<pair<int, int>> v(se.begin(), se.end());
for (int i = 0; i < v.size(); ++i)
for (int j = 0; j <= i; ++j)
mx = max(mx, mp[make_pair(v[i].first + v[j].first, v[i].second + v[j].second)] += i != j ? 2 : 1);
printf("%d", (int)v.size() - mx);
}
#include <bits/stdc++.h>
#define maxm 1000006
using namespace std;
int oc[maxm][30], l1, l2;
char s[maxm], s2[maxm];
int main(void)
{
scanf("%d%d", &l1, &l2);
scanf("%s%s", s + 1, s2 + 1);
for (int i = 0; i < 30; i++)
oc[0][i] = -1;
for (int i = 1; i <= l1; i++)
{
for (int j = 0; j < 26; j++)
{
if (s[l1 - i + 1] - 'a' == j)
oc[i][j] = i;
else
oc[i][j] = oc[i - 1][j];
}
}
int loc = 0, ans = 0, ct = 0;
for (int i = 1; i <= l2; i++)
{
int val = s2[i] - 'a';
for (int j = val + 1; j < 26; j++)
{
if (oc[l1 - loc][j] != -1)
ans = max(ans, i - 1 + oc[l1 - loc][j]);
//cout<<ans<<endl;
}
//if(ans){cout<<ans;return 0;}
if (oc[l1 - loc][val] != -1)
loc = l1 - oc[l1 - loc][val] + 1;
else
{
ct = 1;
break;
}
}
if (ct == 0 && loc != l1)
{
ans = max(ans, l1 - loc + l2);
}
if (ans > 0)
cout << ans;
else
cout << -1;
}