AtCoder Beginner Contest 340（A-G）-编程知识

A - Arithmetic Progression (atcoder.jp)

1.思路：循环输出即可

2.代码：

#include <bits/stdc++.h>
#define rep(i,z,n) for(int i = z;i <= n; i++)
#define per(i,n,z) for(int i = n;i >= z; i--)
#define PII pair<int,int>
#define fi first
#define se second
#define vi vector<int>
#define vl vector<ll>
#define pb push_back
#define sz(x) (int)x.size()
#define all(x) (x).begin(),(x).end()
using namespace std;
using ll = long long;
const int N = 2e5 + 10;
void solve()
{int a,b,c;cin >> a >> b >> c;while (a <= b) {cout << a << ' ';a += c;}
}
int main()
{ios::sync_with_stdio(false);cin.tie(0),cout.tie(0);int T = 1;while(T --) solve();return 0;
}

B - Append (atcoder.jp)

1.思路：vector模拟即可。

2.代码：

#include <bits/stdc++.h>
#define rep(i,z,n) for(int i = z;i <= n; i++)
#define per(i,n,z) for(int i = n;i >= z; i--)
#define PII pair<int,int>
#define fi first
#define se second
#define vi vector<int>
#define vl vector<ll>
#define pb push_back
#define sz(x) (int)x.size()
#define all(x) (x).begin(),(x).end()
using namespace std;
using ll = long long;
const int N = 2e5 + 10;
void solve()
{vector<int> v;int q;cin >> q;while (q --) {int op,x;cin >> op >> x;if (op == 1) {v.pb(x);}else {cout << v[v.size() - x] << '\n';}}
}
int main()
{ios::sync_with_stdio(false);cin.tie(0),cout.tie(0);int T = 1;while(T --) solve();return 0;
}

C - Divide and Divide (atcoder.jp)

1.思路：记忆化搜索，模拟即可。

2.代码：

#include <bits/stdc++.h>
#define rep(i,z,n) for(int i = z;i <= n; i++)
#define per(i,n,z) for(int i = n;i >= z; i--)
#define PII pair<int,int>
#define fi first
#define se second
#define vi vector<int>
#define vl vector<ll>
#define pb push_back
#define sz(x) (int)x.size()
#define all(x) (x).begin(),(x).end()
using namespace std;
using ll = long long;
const int N = 2e5 + 10;
map<ll,ll> mem;
ll dfs(ll x)
{if (x == 1) {return 1ll;}if (mem.count(x)) {return mem[x];}return mem[x] = (dfs(x / 2) + dfs((x + 1) / 2)) + x;
}
void solve()
{ll n;cin >> n;cout << dfs(n / 2) + dfs((n + 1) / 2) << '\n';
}
int main()
{ios::sync_with_stdio(false);cin.tie(0),cout.tie(0);int T = 1;while(T --) solve();return 0;
}

D - Super Takahashi Bros. (atcoder.jp)

1.思路：发现每行给出的实际上可以等价于两条边

一条是i-->i + 1 花费a

一条是i-->c 花费b

2.代码：

#include <bits/stdc++.h>
#define rep(i,z,n) for(int i = z;i <= n; i++)
#define per(i,n,z) for(int i = n;i >= z; i--)
#define PII pair<int,int>
#define fi first
#define se second
#define vi vector<int>
#define vl vector<ll>
#define pb push_back
#define sz(x) (int)x.size()
#define all(x) (x).begin(),(x).end()
using namespace std;
using ll = long long;
const int N = 2e5 + 10;
//从0开始
template<typename T>
class Graph {using pli = pair<T,int>;
private:const int n;vector<vector<pair<int,int>>> g;
public:Graph(int _n): n(_n) {g.resize(n);}void addEdge(int x,int y,int z = 0) {g[x].push_back({y,z});}vector<T> shortestPath(int s) {vector<T> dis(n);memset(dis.data(),0x3f,dis.size() * sizeof(T));priority_queue<pli,vector<pli>,greater<pli>> q;q.push({0,s});dis[s] = 0;while(!q.empty()) {auto [val,ver] = q.top();q.pop();if(val != dis[ver]) {continue;}for(auto &[y,w]: g[ver]) {if(dis[y] > val + w) {dis[y] = val + w;q.push({dis[y],y});}}}return dis;}
};
void solve()
{int n;cin >> n;Graph<ll> g(n);for (int i = 1;i < n;i ++) {int a,b,c;cin >> a >> b >> c;g.addEdge(i - 1,i,a);g.addEdge(i - 1,c - 1,b);        }auto p = g.shortestPath(0);cout << p[n - 1];
}
int main()
{ios::sync_with_stdio(false);cin.tie(0),cout.tie(0);int T = 1;while(T --) solve();return 0;
}

E - Mancala 2 (atcoder.jp)

1.思路：按照模拟的过程可以分为三段去操作，从当前位置到末尾+轮多少圈+剩下的次数。

发现这是一个需要维护区间加和单点查询的数据结构题目，因此我们考虑直接用树状数组维护原数组差分即可。或者直接用线段树维护原数组也行。

2.代码：

#include <bits/stdc++.h>
#define rep(i,z,n) for(int i = z;i <= n; i++)
#define per(i,n,z) for(int i = n;i >= z; i--)
#define PII pair<int,int>
#define fi first
#define se second
#define vi vector<int>
#define vl vector<ll>
#define pb push_back
#define sz(x) (int)x.size()
#define all(x) (x).begin(),(x).end()
using namespace std;
using ll = long long;
const int N = 2e5 + 10;
template <typename T>
struct Fenwick {const int n;std::vector<T> a;Fenwick (int n) : n(n), a(n + 1) {}void add(int pos, T x) {for (int i = pos; i <= n; i += i & -i) {a[i] += x;}}T query(int x) {T res = 0;for (int i = x; i; i -= i & -i) {res += a[i];}return res;}T query(int l, int r) {if (l == 0 || l > r) {return 0;}return query(r) - query(l - 1);}//找到大于k得第一个地方T kth(int k) {int pos = 0;for(int j = 31 - __builtin_clz(n);j >= 0;j --) {if(pos + (1 << j) <= n && k > a[pos + (1 << j)]) {pos += 1 << j;k -= a[pos];}}return pos + 1;}
};
void solve()
{int n,q;cin >> n >> q;Fenwick<ll> fen(n);int pre = 0,x;for (int i = 1;i <= n;i ++) {cin >> x;fen.add(i,x - pre);pre = x;}for (int i = 1;i <= q;i ++) {cin >> x;x ++;ll rs = fen.query(x);ll mi = min(rs,0ll + n - x);fen.add(x,-rs);fen.add(x + 1,rs);if (mi) {rs -= mi;fen.add(x + 1,1);fen.add(x + 1 + mi,-1);}if (rs) {ll cz = rs / n,cm = rs % n;fen.add(1,cz);fen.add(1,1);fen.add(1 + cm,-1);}}for (int i = 1;i <= n;i ++) {cout << fen.query(i) << ' ';}
}
int main()
{ios::sync_with_stdio(false);cin.tie(0),cout.tie(0);int T = 1;while(T --) solve();return 0;
}

F - S = 1 (atcoder.jp)

1.思路：考虑使用鞋带公式化简，我们可以得到，这么个方程|ay - bx| = 2，考虑把绝对值拆开，做两次exgcd求解即可。

2.代码：

#include <bits/stdc++.h>
#define rep(i,z,n) for(int i = z;i <= n; i++)
#define per(i,n,z) for(int i = n;i >= z; i--)
#define PII pair<int,int>
#define fi first
#define se second
#define vi vector<int>
#define vl vector<ll>
#define pb push_back
#define sz(x) (int)x.size()
#define all(x) (x).begin(),(x).end()
using namespace std;
using ll = long long;
const int N = 2e5 + 10;
ll exgcd(ll a, ll b, ll &x, ll &y)
{if (!b){x = 1, y = 0;return a;}ll d = exgcd(b, a % b, y, x);y -= a / b * x;return d;
}
void solve()
{ll a,b;cin >> a >> b;ll x,y;ll d1 = exgcd(a,-b,y,x);if (2 % d1) {ll d2 = exgcd(b,-a,x,y);x *= 2 / d2,y *= 2 / d2;if (2 % d2) {cout << -1 << endl;}else {cout << x << ' ' << y << '\n';}}else {x *= 2 / d1,y *= 2 / d1;cout << x << ' ' << y << '\n';}
}
int main()
{ios::sync_with_stdio(false);cin.tie(0),cout.tie(0);int T = 1;while(T --) solve();return 0;
}

G - Leaf Color (atcoder.jp)

1.思路：考虑每个遍历每个颜色单独求一遍答案，因此可以建一棵虚树进行dp即可。

f[i] 表示以i为根节点，（不管符不符合的答案为多少），g[i]表示 $\sum_{j \epsilon son[i]}f[j]$ 。

当i为枚举的颜色时，答案加上f[i]的方案数即可。

当i不是枚举的颜色时，首先f[i]应该-1（因为i不能作为根节点），其次答案应该加上f[i] - g[i]（所有选择一个子树的方案和）。

2.代码：

#include <bits/stdc++.h>
#define rep(i,z,n) for(int i = z;i <= n; i++)
#define per(i,n,z) for(int i = n;i >= z; i--)
#define PII pair<int,int>
#define fi first
#define se second
#define vi vector<int>
#define vl vector<ll>
#define pb push_back
#define sz(x) (int)x.size()
#define all(x) (x).begin(),(x).end()
using namespace std;
using ll = long long;
const int N = 2e5 + 10;
using i64 = long long;constexpr int P = 998244353;
// assume -P <= x < 2P
int Vnorm(int x) {if (x < 0) {x += P;}if (x >= P) {x -= P;}return x;
}
template<class T>
T power(T a, i64 b) {T res = 1;for (; b; b /= 2, a *= a) {if (b % 2) {res *= a;}}return res;
}
struct Mint {int x;Mint(int x = 0) : x(Vnorm(x)) {}Mint(i64 x) : x(Vnorm(x % P)) {}int val() const {return x;}Mint operator-() const {return Mint(Vnorm(P - x));}Mint inv() const {assert(x != 0);return power(*this, P - 2);}Mint &operator*=(const Mint &rhs) {x = i64(x) * rhs.x % P;return *this;}Mint &operator+=(const Mint &rhs) {x = Vnorm(x + rhs.x);return *this;}Mint &operator-=(const Mint &rhs) {x = Vnorm(x - rhs.x);return *this;}Mint &operator/=(const Mint &rhs) {return *this *= rhs.inv();}friend Mint operator*(const Mint &lhs, const Mint &rhs) {Mint res = lhs;res *= rhs;return res;}friend Mint operator+(const Mint &lhs, const Mint &rhs) {Mint res = lhs;res += rhs;return res;}friend Mint operator-(const Mint &lhs, const Mint &rhs) {Mint res = lhs;res -= rhs;return res;}friend Mint operator/(const Mint &lhs, const Mint &rhs) {Mint res = lhs;res /= rhs;return res;}friend std::istream &operator>>(std::istream &is, Mint &a) {i64 v;is >> v;a = Mint(v);return is;}friend std::ostream &operator<<(std::ostream &os, const Mint &a) {return os << a.val();}
};
vector<int> son[N],g[N];
int c[N],dep[N],dfn[N],fa[N][20],stk[N],top,tim;
inline void dfs(int u,int f)
{dfn[u] = ++ tim;dep[u] = dep[f] + 1;fa[u][0] = f;for(int i = 1;i <= 18;i ++) fa[u][i] = fa[fa[u][i - 1]][i - 1];for(auto v : g[u]) {if(v == f) continue;dfs(v,u);}
}
inline int LCA(int x,int y)
{if(dep[x] > dep[y]) swap(x,y);int d = dep[y] - dep[x];for(int i = 18;i >= 0;i --) {if(d >> i & 1) y = fa[y][i];}if(x == y) return x;for(int i = 18;i >= 0;i --) {if(fa[x][i] != fa[y][i]) {x = fa[x][i];y = fa[y][i];}}return fa[x][0];
}
Mint ans = 0;
int tarc = 0;
void add(int &a,int &b)
{son[a].push_back(b);
}
Mint DP(int x)
{Mint f = 1,g = 0;for (auto &y: son[x]) {auto nf = DP(y);f *= (nf + 1);g += nf;}if (c[x] == tarc) {ans += f;}else {f -= 1;ans += f - g;}son[x].clear();return f;
}
void buildTree(vector<int> &ver)
{if (ver.size() == 1) {ans += 1;return;}sort(ver.begin(),ver.end(),[&](int &x,int &y){return dfn[x] < dfn[y];});stk[top = 1] = 1;if (ver[0] != 1) {stk[++ top] = ver[0];}for (int i = 1;i < ver.size();i ++) {int lca = LCA(ver[i],stk[top]);while (top > 1 && dfn[stk[top - 1]] >= dfn[lca]) {add(stk[top - 1],stk[top]);top --;}if (lca != stk[top]) {add(lca,stk[top]);stk[top] = lca;}stk[++ top] = ver[i];}while (top) {add(stk[top - 1],stk[top]);top --;}DP(1);
}
vector<int> cor[N];
void solve()
{int n;cin >> n;for (int i = 1;i <= n;i ++) {cin >> c[i];cor[c[i]].push_back(i);}for (int i = 1;i < n;i ++) {int a,b;cin >> a >> b;g[a].push_back(b);g[b].push_back(a);}dfs(1,0);for (int i = 1;i <= n;i ++) {if (cor[i].size()) {tarc = i;buildTree(cor[i]);}}cout << ans;
}
int main()
{ios::sync_with_stdio(false);cin.tie(0),cout.tie(0);int T = 1;while(T --) solve();return 0;
}