ARC刷题笔记
arc064
[ARC064E] Cosmic Rays
建图 跑dijkstra即可
// Author: xiaruize
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define mms(arr, n) memset(arr, n, sizeof(arr))
#define rep(i, a, n) for (int i = (a); i <= (n); ++i)
#define per(i, n, a) for (int i = (n); i >= (a); --i)
int max(int a, int b)
{
if (a > b)
return a;
return b;
}
int min(int a, int b)
{
if (a < b)
return a;
return b;
}
const int INF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1000000007;
const int N = 1e3 + 10;
// bool st;
struct node
{
int x, y, r;
} s[N];
int n;
vector<pair<int, double>> g[N];
double dis[N];
// bool en;
void solve()
{
cin >> s[0].x >> s[0].y >> s[1].x >> s[1].y;
s[0].r = s[1].r = 0;
cin >> n;
n++;
rep(i, 2, n) cin >> s[i].x >> s[i].y >> s[i].r;
rep(i, 0, n)
{
dis[i] = INF;
rep(j, 0, n)
{
if (i == j)
continue;
g[i].push_back({j, max((double)0, sqrt((s[i].x - s[j].x) * (s[i].x - s[j].x) + (s[i].y - s[j].y) * (s[i].y - s[j].y)) - s[i].r - s[j].r)});
}
}
dis[0] = 0;
priority_queue<pair<double, int>, vector<pair<double, int>>, greater<pair<double, int>>> q;
q.push({0, 0});
while (!q.empty())
{
auto [ds, x] = q.top();
q.pop();
if (dis[x] != ds)
continue;
if (x == 1)
{
cout << fixed << setprecision(10) << ds << endl;
return;
}
for (auto [v, w] : g[x])
{
if (dis[v] > dis[x] + w)
{
dis[v] = dis[x] + w;
q.push({dis[v], v});
}
}
}
}
signed main()
{
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
// cerr<<(&en-&st)/1024.0/1024.0<<endl;
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int testcase = 1;
// cin >> testcase;
while (testcase--)
solve();
return 0;
}
[ARC064F] Rotated Palindromes
因为原串是回文串，所以它的循环节必然为回文串
考虑枚举最小的循环节的大小
设长度为 \(x\) 的最小循环节的可能数 \(f(x)\)
那么
\[f(x)=k^{\lceil\frac{x}{2}\rceil}-\sum\limits _{v|x} f(v)
\]
- 对于每个长度 \(x\) 为奇数的循环节，在 \(2\) 操作后会有 \(x\) 种不同情况
- 对于每个长度 \(x\) 为偶数的循环节，在 \(2\) 操作后会有 \(\frac{x}{2}\) 种不同情况
累加贡献即可
// Author: xiaruize
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define mms(arr, n) memset(arr, n, sizeof(arr))
#define rep(i, a, n) for (int i = (a); i <= (n); ++i)
#define per(i, n, a) for (int i = (n); i >= (a); --i)
int max(int a, int b)
{
if (a > b)
return a;
return b;
}
int min(int a, int b)
{
if (a < b)
return a;
return b;
}
const int INF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1000000007;
const int N = 2e5 + 10;
int qpow(int a, int b)
{
int res = 1;
while (b)
{
if (b & 1)
res = res * a % MOD;
a = a * a % MOD;
b >>= 1;
}
return res;
}
// bool st;
int n, m;
int res = 0;
vector<int> vec;
map<int, int> dp;
// bool en;
void solve()
{
cin >> n >> m;
rep(i, 1, n)
{
if (i * i > n)
break;
if (n % i == 0)
{
vec.push_back(i);
if (i * i != n)
vec.push_back(n / i);
}
}
sort(ALL(vec));
for (auto v : vec)
{
dp[v] = qpow(m, (v + 1) / 2);
for (auto p : vec)
{
if (p >= v)
break;
if (v % p == 0)
dp[v] = ((dp[v] - dp[p]) % MOD + MOD) % MOD;
}
// cerr << dp[v] << ' ';
}
for (auto v : dp)
{
if (v.first % 2 == 1)
res = (res + v.second * v.first % MOD) % MOD;
else
res = (res + v.second * v.first / 2 % MOD) % MOD;
}
cout << res << endl;
}
signed main()
{
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
// cerr<<(&en-&st)/1024.0/1024.0<<endl;
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int testcase = 1;
// cin >> testcase;
while (testcase--)
solve();
return 0;
}
arc065
[ARC065F] シャッフル
\(dp_{i,j}\) 表示考虑到 \(i\) 用了 \(j\) 个 \(1\) 的方案数
考虑转移的时候，不能从 \([1,i-1]\) 转移，因为有一些是不可能取到的
预处理出前缀和 \(sum_i\)，当前点能交换到的最右边的位置 \(r_i\)
则 \(i\) 这个位置能被转移到的区间为 \([\max{(0,sum_{r_i}-r_i+i)},\min{(i,sum_{r_i})}]\)
直接转移即可
\[dp_{i,j}\leftarrow \begin{cases}
dp_{i-1,j}\\
dp_{i-1,j-1} (j\neq 0)
\end{cases}
\]
// Author: xiaruize
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define mms(arr, n) memset(arr, n, sizeof(arr))
#define rep(i, a, n) for (int i = (a); i <= (n); ++i)
#define per(i, n, a) for (int i = (n); i >= (a); --i)
int max(int a, int b)
{
if (a > b)
return a;
return b;
}
int min(int a, int b)
{
if (a < b)
return a;
return b;
}
const int INF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1000000007;
const int N = 3e3 + 10;
// bool st;
int n, m;
char s[N];
int sum[N];
int r[N];
int dp[N][N];
// bool en;
void solve()
{
cin >> n >> m;
cin >> (s + 1);
rep(i, 1, n)
{
sum[i] = sum[i - 1] + (s[i] == '1');
}
rep(i, 1, m)
{
int st, en;
cin >> st >> en;
r[st] = max(r[st], en);
}
dp[0][0] = 1;
rep(i, 1, n) r[i] = max(r[i], max(r[i - 1], i));
rep(i, 1, n)
{
rep(j, max(0, sum[r[i]] - (r[i] - i)), min(i, sum[r[i]]))
{
dp[i][j] += dp[i - 1][j];
if (j)
(dp[i][j] += dp[i - 1][j - 1]) %= MOD;
}
}
cout << dp[n][sum[n]] << endl;
}
signed main()
{
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
// cerr<<(&en-&st)/1024.0/1024.0<<endl;
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int testcase = 1;
// cin >> testcase;
while (testcase--)
solve();
return 0;
}
[ARC065E] へんなコンパス
两点之间的距离是恒定不变的
考虑将曼哈顿距离转换为切比雪夫距离，具体来说就是 \((x,y)\rightarrow (x+y,x-y)\) 此时两个点 \((x_1,y_1),(x_2,y_2)\) 之间的距离为 \(\max(|x_1-x_2|,|y_1-y_2|)\)
设原来的距离为 \(dis\)， 则距离为 \(dis\) 的两点需要满足
\[
\begin{cases}
|x_1-x_2|=dis\\
|y_1-y_1|\leq dis
\end{cases}
或
\begin{cases}
|x_1-x_2|\leq dis\\
|y_1-y_1|= dis
\end{cases}
\]
考虑如何计算满足第一个条件的点对数量，可以按 \(x\) 排序，然后对于每个点，合法区间的左右端点都具有单调性，two-pointers即可
然后将 \((x,y)\rightarrow (y,x)\) 再做上述操作
但是这样会在 \(|x_1-x_2|=|y_1-y_2|\) 时计算重复，所以第二次时保证 \(|y_1-y_2|<dis\) 即可
感觉luogu黑题有点虚标，可能你看到这个的时候已经不是了吧~~~
// Author: xiaruize
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define mms(arr, n) memset(arr, n, sizeof(arr))
#define rep(i, a, n) for (int i = (a); i <= (n); ++i)
#define per(i, n, a) for (int i = (n); i >= (a); --i)
int max(int a, int b)
{
if (a > b)
return a;
return b;
}
int min(int a, int b)
{
if (a < b)
return a;
return b;
}
const int INF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1000000007;
const int N = 2e5 + 10;
// bool st;
struct node
{
int x, y, id;
} s[N];
int n, a, b;
struct union_set
{
int fa[N], sz[N];
void init()
{
rep(i, 1, n) fa[i] = i, sz[i] = 0;
}
int get(int x)
{
if (fa[x] == x)
return x;
return fa[x] = get(fa[x]);
}
void merge(int x, int y)
{
int fx = get(x), fy = get(y);
if (fx == fy)
return;
fa[fy] = fx;
sz[fx] += sz[fy];
sz[fy] = 0;
}
} uni;
// bool en;
bool cmp(node a, node b)
{
if (a.x != b.x)
return a.x < b.x;
return a.y < b.y;
}
void calc(int dis, int cur)
{
int l, r;
l = r = 1;
int la = 1;
rep(i, 1, n)
{
while (l <= n && s[i].x - s[l].x > dis)
l++;
while (l <= n && s[i].x - s[l].x == dis && s[i].y - s[l].y > dis - cur)
l++;
while (r <= n && s[i].x - s[r].x > dis)
r++;
while (r <= n && s[i].x - s[r].x == dis && s[r].y - s[i].y <= dis - cur)
r++;
if (l >= r || s[i].x - s[l].x != dis)
continue;
// cerr << i << ' ' << l << ' ' << r - 1 << endl;
uni.sz[uni.get(s[i].id)] += r - l;
la = max(la, l);
if (la >= r)
continue;
uni.merge(s[i].id, s[la].id);
while (la < r - 1)
{
uni.merge(s[la].id, s[la + 1].id);
la++;
}
}
}
void solve()
{
cin >> n >> a >> b;
uni.init();
rep(i, 1, n)
{
int x, y;
cin >> x >> y;
s[i].x = x + y;
s[i].y = x - y;
s[i].id = i;
}
int dis = max(abs(s[a].x - s[b].x), abs(s[a].y - s[b].y));
sort(s + 1, s + n + 1, cmp);
calc(dis, 0);
rep(i, 1, n) swap(s[i].x, s[i].y);
sort(s + 1, s + n + 1, cmp);
calc(dis, 1);
cout << uni.sz[uni.get(a)] << endl;
}
signed main()
{
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
// cerr<<(&en-&st)/1024.0/1024.0<<endl;
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int testcase = 1;
// cin >> testcase;
while (testcase--)
solve();
return 0;
}
arc092
[ABC091D] Two Sequences
考虑对于每一位计算贡献
\(a_i+b_j\) 的 \(2^k\) 为 \(1\) 当且仅当 \(2^k\leq a_i+b_j<2^{k+1}\) 或 \(3\times 2^k\leq a_i+b_j<4\times 2^{k+1}\)
对于每一位，取模后对排序，二分即可
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