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如果这篇博客帮助到你，可以请我喝一杯咖啡~
最后一分钟提交不上去，不然 E 就过了呜呜呜。
Telephone Number
#include <bits/stdc++.h>
using namespace std;
const int N = 127;
char s[N];
int t, n;
int main()
{
for (scanf("%d", &t); t--;)
{
scanf("%d%s", &n, s);
printf(find(s, s + n, '8') > s + n - 11 ? "NO\n" : "YES\n");
}
}
Lost Numbers
暴力 check 一下即可。
#include <bits/stdc++.h>
using namespace std;
int a[6] = {4, 8, 15, 16, 23, 42}, b[6];
int main()
{
for (int i = 0; i < 4; ++i)
{
printf("? %d %d\n", i + 1, i + 2), fflush(stdout);
scanf("%d", &b[i]);
}
do
if (a[0] * a[1] == b[0] && a[1] * a[2] == b[1] && a[2] * a[3] == b[2] && a[3] * a[4] == b[3])
return printf("! %d %d %d %d %d %d", a[0], a[1], a[2], a[3], a[4], a[5]), 0;
while (next_permutation(a, a + 6));
}
News Distribution
每个 group 里的人都是连通的，最后求一下每个人所在连通块的大小即可。使用并查集实现。
#include <bits/stdc++.h>
using namespace std;
struct UnionfindSet : vector<int>
{
vector<int> siz;
UnionfindSet(int n) : vector<int>(n), siz(n, 1)
{
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)
siz[at(w) = u] += siz[w];
}
int ask(int u) { return at(u) != u ? at(u) = ask(at(u)) : u; }
};
int main()
{
int n, m;
scanf("%d%d", &n, &m);
UnionfindSet ufs(n + 1);
for (int i = 0, w, u, k; i < m; ++i)
if (scanf("%d", &k), k)
for (scanf("%d", &u); --k; ufs.merge(u, w))
scanf("%d", &w);
for (int i = 1; i <= n; ++i)
printf("%d ", ufs.siz[ufs.ask(i)]);
}
Bicolored RBS
贪心。
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 9;
char s[N];
int n;
int main()
{
scanf("%d%s", &n, s);
for (int i = 0, d[2] = {0, 0}; i < n; ++i)
{
if (s[i] == '(')
putchar('0' + (d[0] > d[1])), ++d[d[0] > d[1]];
else
putchar('0' + (d[0] < d[1])), --d[d[0] < d[1]];
}
}
比赛时候的二分解法
比赛的时候 sb 了，敲了一个二分。
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 9;
char s[N], ans[N];
int n;
int work(int m)
{
for (int i = 0, d[2] = {0, 0}; i < n; ++i)
{
if (s[i] == '(')
{
ans[i] = '0' + (d[0] > d[1]);
if (++d[d[0] > d[1]] > m)
return 0;
}
else
ans[i] = '0' + (d[0] < d[1]), --d[d[0] < d[1]];
}
return 1;
}
int bs(int b, int e)
{
if (e - b < 2)
return e;
int m = b + e >> 1;
return work(m) ? bs(b, m) : bs(m, e);
}
int main()
{
scanf("%d%s", &n, s);
work(bs(0, n));
printf("%s", ans);
}
Range Deleting
手速慢了一丢丢，最后一分钟没交上去，赛后马上 ac…😔
枚举$1,2,\dots,x$里的所有上界$r$，找到对应最大的下界$l$使得$[l,r]$是符合要求，把$l$加入答案即可。比赛的时候这个下界是二分去找的，然而结束之后我仔细想了一下，这个下界$l$是随着$r$单调的，因此可以直接用双指针去维护，于是得到了下面这个线性的做法。
理想是美好的，然而双指针的线性做法居然比二分的做法运行时间要长也是醉了。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N = 1e6 + 9;
ll n, x, y, ans, mi[N], ma[N], mma[N];
int main()
{
scanf("%lld%lld", &n, &x);
for (ll i = 1, a; i <= n; ++i)
{
scanf("%lld", &a);
if (!mi[a])
mi[a] = i;
ma[a] = i;
}
y = x;
for (ll i = 1; i <= x; ++i)
{
mma[i] = max(mma[i - 1], ma[i]);
if (y == x && mi[i] && mi[i] < mma[i - 1])
y = i;
}
for (; x; --x)
{
while (y > x || mma[y - 1] > n)
--y;
ans += y;
if (ma[x] > n)
break;
if (mi[x])
n = min(n, mi[x]);
}
printf("%lld", ans);
}
比赛时候的二分解法
为啥我比赛的时候老想着二分…
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N = 1e6 + 9;
ll n, x, se, ans, mi[N], ma[N], bmi[N], bma[N];
ll bs(ll b, ll e, ll pos)
{
if (e - b < 2)
return b;
ll m = b + e >> 1;
return bmi[m] > pos ? bs(b, m, pos) : bs(m, e, pos);
}
int main()
{
scanf("%lld%lld", &n, &x);
for (int i = 1, a; i <= n; ++i)
{
scanf("%d", &a);
if (!mi[a])
mi[a] = ma[a] = i;
else
ma[a] = i;
}
bmi[x + 1] = n + 1;
for (ll i = x; i; --i)
{
bmi[i] = bmi[i + 1];
if (mi[i])
bmi[i] = min(bmi[i], mi[i]), bma[i] = max(bma[i + 1], ma[i]);
if (!se && ma[i] && ma[i] > bmi[i + 1])
se = i;
}
for (ll i = 1, mma = 0; i <= x; ++i)
{
ans += x - bs(max(i, se), x + 1, mma) + 1;
if (mi[i] && mi[i] < mma)
break;
mma = max(mma, ma[i]);
}
printf("%lld", ans);
}
Scalar Queries
从小到大枚举 a，计算其对答案的贡献。
用两个树状数组分别维护已经考虑过的数里下标前 w 小和前 k 大的。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 5e5 + 7;
struct Mod
{
const ll M;
Mod(ll M) : M(M) {}
ll qadd(ll a, ll b) const { return a += b, a < M ? a : a - M; }
ll add(ll a, ll b) const { return qadd((a + b) % M, M); }
} M(1e9 + 7);
struct BaseFenwick : vector<ll>
{
BaseFenwick(int n) : vector<ll>(n, 0) {}
void add(int x, ll w)
{
for (; x < size(); x += x & -x)
at(x) = M.qadd(at(x), w);
}
ll ask(int x)
{
ll ans = 0;
for (; x; x -= x & -x)
ans = M.qadd(ans, at(x));
return ans;
}
} t[2] = {N, N};
pair<int, int> a[N];
int n, ans;
int main()
{
scanf("%d", &n);
for (int i = 0; i < n; ++i)
scanf("%d", &a[i].first), a[i].second = i;
sort(a, a + n);
for (int i = 0, w, k; i < n; ++i)
{
t[1].add(k = n - a[i].second, w = a[i].second + 1);
t[0].add(k, k);
ans = M.add(ans, a[i].first * M.add(w * t[0].ask(k), k * M.add(t[1].ask(n), -t[1].ask(k))));
}
printf("%d", ans);
}
Low Budget Inception
怪长的题面，大意是计算「盗梦空间」中扭曲空间的角度。按题意找到，双指针强力模拟即可。
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 9;
double PI = acos(-1), ans;
int n, d, m, x, cur, dl, dr, a[N];
int main()
{
scanf("%d%d", &n, &d);
for (int i = 1; i <= n; ++i)
scanf("%d", &a[i]);
for (scanf("%d", &m); m--; printf("%.9lf\n", ans))
{
for (scanf("%d", &x); cur < n && a[cur + 1] < x;)
++cur;
if (a[cur] == x - 1 && a[cur + 1] == x)
ans = PI;
else if (a[cur] == x - 1 || a[cur + 1] == x)
ans = PI / 2;
else
{
ans = 0;
if (cur)
ans = max(ans, atan2(1, dl = x - a[cur] - 1));
if (cur < n)
ans = max(ans, atan2(1, dr = a[cur + 1] - x));
for (int pl = cur, pr = cur + 1, d = min(dl, dr) * 4 + 20; pl && pr <= n && a[pr] - a[pl] <= d; dl < dr ? --pl : ++pr)
{
dl = x - a[pl] - 1, dr = a[pr] - x;
if (abs(dl - dr) < 2)
{
ans = max(ans, 2 * atan2(1, max(dl, dr)));
break;
}
}
}
}
}