线上算法赛第一次4道题AC，是啥感觉？
碎碎念
本次线上赛有很多值得纪念的
- 最简单的A题竟然提交了2次，又是被老码农一顿奚落(⊙︿⊙)；
- 第一次30分钟内解决了A、B、D三道题，撒花✿✿ヽ(°▽°)ノ✿；
- B、C、D三道题一次提交AC，呱唧呱唧(#^.^#)
C题其实也不难，结果一开始题就理解错了，思考的方向跑偏了，o(╥﹏╥)o 思路：map结合优先队列去做，后来发现思路行不通，所以耽误了不少时间。
期间还看了E、F两道题描述（E、F题解做了整理，以后在学习）
- E是动态规划相关的，学艺不精，唯有继续努力，冲鸭！
- F题赛后看题解需要用到树状数组，知识盲区，继续加油(*￣︶￣)
ABC253-01
走过路过不要错过，怪事速报来了||ヽ(￣▽￣)ノミ|Ю
- 本次大赛好几位大神都在E题提交了2次，不知道为何会同时出现这种事。正常大神都是一次AC
- 老码农嘟嘟囔囔说：你这小孩见不得别人好，离大神的功力还十万步千里呢。。。
- tourist大神一出江湖，基本上都是第一名，得第2名都是偶尔，望尘莫及，唯有努力学习，٩(๑>◡<๑)۶
A - Median?
- https://atcoder.jp/contests/abc253/tasks/abc253_a
小码匠题解
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
using vll = vector<ll>;
#define rep(n) for(int i = 0; i < n; i++)
#define f(i, start, end) for(int i = start; i < end; i++)
#define per(i, n) for(int i = n - 1; i >= 0; i--)
#define all(x) begin(x), end(x)
#define sort_all(a) sort(a.begin, a.end)
#define fi first
#define se second
#define endl '\n'
#define out(x) cout << x << '\n'
int main() {
// 提升cin、cout效率
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int a, b, c;
cin >> a >> b >> c;
if((a + b + c) - min(a, min(b, c)) - max(a, max(b, c)) == b) {
cout << "Yes";
} else {
cout << "No";
}
return 0;
}
参考题解：膜拜大神
#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int a, b, c;
cin >> a >> b >> c;
vector<int> v = {a, b, c};
sort(v.begin(), v.end());
cout << (b == v[1] ? "Yes" : "No") << '\n';
return 0;
}
B - Distance Between Token
- https://atcoder.jp/contests/abc253/tasks/abc253_b
小码匠题解
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
using vll = vector<ll>;
#define rep(n) for(int i = 0; i < n; i++)
#define f(i, start, end) for(int i = start; i < end; i++)
#define per(i, n) for(int i = n - 1; i >= 0; i--)
#define all(x) begin(x), end(x)
#define sort_all(a) sort(a.begin, a.end)
#define fi first
#define se second
#define endl '\n'
#define out(x) cout << x << '\n'
int main() {
// 提升cin、cout效率
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
// 输入
int n, w;
cin >> n >> w;
pair<int, int> a;
string s;
int ans = 0;
for(int i = 0; i < n; i++) {
cin >> s;
for(int j = 0; j < w; j++) {
if(s[j] == 'o'){
if(ans == 0) {
a.fi = i;
a.se = j;
ans++;
} else {
cout << abs(i - a.fi) + abs(j - a.se);
return 0;
}
}
}
}
}
参考题解（膜拜大神）
#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int h, w;
cin >> h >> w;
vector<string> s(h);
for (int i = 0; i < h; i++) {
cin >> s[i];
}
int xa = -1, ya = -1, xb = -1, yb = -1;
for (int i = 0; i < h; i++) {
for (int j = 0; j < w; j++) {
if (s[i][j] == 'o') {
if (xa == -1) {
xa = i;
ya = j;
} else {
xb = i;
yb = j;
}
}
}
}
cout << abs(xa - xb) + abs(ya - yb) << '\n';
return 0;
}
C - Max - Min Query
- https://atcoder.jp/contests/abc253/tasks/abc253_c
小码匠题解
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
using vll = vector<ll>;
#define rep(n) for(int i = 0; i < n; i++)
#define f(i, start, end) for(int i = start; i < end; i++)
#define per(i, n) for(int i = n - 1; i >= 0; i--)
#define all(x) begin(x), end(x)
#define sort_all(a) sort(a.begin, a.end)
#define fi first
#define se second
#define endl '\n'
#define out(x) cout << x << '\n'
int main() {
// 提升cin、cout效率
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
long long q;
cin >> q;
int a, b, c;
set<int> query;
unordered_map<int, int> temp;
for(int i = 0; i < q; i++) {
cin >> a;
if(a == 1) {
cin >> b;
query.insert(b);
temp[b]++;
} else if(a == 2) {
cin >> b >> c;
temp[b] -= min(temp[b], c);
if(temp[b] == 0) {
query.erase(b);
}
} else if(a == 3) {
cout << *query.rbegin() - *query.begin() << endl;
}
}
return 0;
}
参考题解（膜拜大神）
#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int q;
cin >> q;
multiset<int> s;
while (q--) {
int op;
cin >> op;
if (op == 1) {
int x;
cin >> x;
s.insert(x);
}
if (op == 2) {
int x, c;
cin >> x >> c;
while (c--) {
auto it = s.find(x);
if (it == s.end()) {
break;
}
s.erase(it);
}
}
if (op == 3) {
cout << (*prev(s.end()) - *s.begin()) << '\n';
}
}
return 0;
}
D - FizzBuzz Sum Hard
- https://atcoder.jp/contests/abc253/tasks/abc253_d
小码匠题解
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
using vll = vector<ll>;
#define rep(i, n) for(int i = 0; i < n; i++)
#define f(i, start, end) for(int i = start; i < end; i++)
#define per(i, n) for(int i = n - 1; i >= 0; i--)
#define all(x) begin(x), end(x)
#define sort_all(a) sort(a.begin, a.end)
#define fi first
#define se second
#define endl '\n'
#define out(x) cout << x << '\n'
int main() {
// 提升cin、cout效率
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
long long n, a, b, c;
cin >> n >> a >> b;
c = a / __gcd(a, b) * b;
long long d, e, f;
d = n / c;
e = n / a;
f = n / b;
cout << (1 + n) * n / 2 - (e * a + a) * e / 2 - (f * b + b) * f / 2 + (d * c + c) * d / 2;
return 0;
}
参考题解（膜拜大神）
#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n, a, b;
cin >> n >> a >> b;
auto Sum = [&](long long x) {
return x * (x + 1) / 2;
};
long long c = a / __gcd(a, b) * 1LL * b;
cout << Sum(n) - Sum(n / a) * a - Sum(n / b) * b + Sum(n / c) * c << '\n';
return 0;
}
E - Distance Sequence
- https://atcoder.jp/contests/abc253/tasks/abc253_e
参考题解（官方提供）
動的計画法で解きます。
dp[i][j] を、A の先頭から i 項を決めて、i 項目が j であるような場合の数とします。この dp の遷移は
dp[i+1][j]=(dp[i][1]+…+dp[i][j−K])+(dp[i][j+K]+…+dp[i][M])
です。（なお、1>j−K のときやj+K>M のとき、K=0 のときは微妙に遷移が異なるので注意してください。）この dp では状態数が O(NM) 、遷移が O(M) となり、全体の計算量は O(NM
2
) となり実行時間制限に間に合いません。
ここで、dp[i+1] への遷移を考える際に事前に dp[i] の累積和を求めておくことで、遷移が O(1) で可能になり、全体の計算量が O(NM) になります。これは十分高速です。
- 参考代码
#include <bits/stdc++.h>
using namespace std;
const long long MOD = 998244353;
int main(){
int N, M, K;
cin >> N >> M >> K;
vector<vector<long long>> dp(N, vector<long long>(M, 0));
for (int i = 0; i < M; i++){
dp[0][i] = 1;
}
for (int i = 0; i < N - 1; i++){
vector<long long> S(M + 1);
S[0] = 0;
for (int j = 0; j < M; j++){
S[j + 1] = S[j] + dp[i][j];
S[j + 1] %= MOD;
}
for (int j = 0; j < M; j++){
dp[i + 1][j] = S[M];
if (K > 0){
dp[i + 1][j] -= S[min(j + K, M)] - S[max(j - K + 1, 0)];
}
dp[i + 1][j] += MOD;
dp[i + 1][j] %= MOD;
}
}
long long ans = 0;
for (int i = 0; i < M; i++){
ans += dp[N - 1][i];
}
ans %= MOD;
cout << ans << endl;
}
参考题解（膜拜大神）
#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif
template <typename T>
T inverse(T a, T m) {
T u = 0, v = 1;
while (a != 0) {
T t = m / a;
m -= t * a; swap(a, m);
u -= t * v; swap(u, v);
}
assert(m == 1);
return u;
}
template <typename T>
class Modular {
public:
using Type = typename decay<decltype(T::value)>::type;
constexpr Modular() : value() {}
template <typename U>
Modular(const U& x) {
value = normalize(x);
}
template <typename U>
static Type normalize(const U& x) {
Type v;
if (-mod() <= x && x < mod()) v = static_cast<Type>(x);
else v = static_cast<Type>(x % mod());
if (v < 0) v += mod();
return v;
}
const Type& operator()() const { return value; }
template <typename U>
explicit operator U() const { return static_cast<U>(value); }
constexpr static Type mod() { return T::value; }
Modular& operator+=(const Modular& other) { if ((value += other.value) >= mod()) value -= mod(); return *this; }
Modular& operator-=(const Modular& other) { if ((value -= other.value) < 0) value += mod(); return *this; }
template <typename U> Modular& operator+=(const U& other) { return *this += Modular(other); }
template <typename U> Modular& operator-=(const U& other) { return *this -= Modular(other); }
Modular& operator++() { return *this += 1; }
Modular& operator--() { return *this -= 1; }
Modular operator++(int) { Modular result(*this); *this += 1; return result; }
Modular operator--(int) { Modular result(*this); *this -= 1; return result; }
Modular operator-() const { return Modular(-value); }
template <typename U = T>
typename enable_if<is_same<typename Modular<U>::Type, int>::value, Modular>::type& operator*=(const Modular& rhs) {
#ifdef _WIN32
uint64_t x = static_cast<int64_t>(value) * static_cast<int64_t>(rhs.value);
uint32_t xh = static_cast<uint32_t>(x >> 32), xl = static_cast<uint32_t>(x), d, m;
asm(
"divl %4; \n\t"
: "=a" (d), "=d" (m)
: "d" (xh), "a" (xl), "r" (mod())
);
value = m;
#else
value = normalize(static_cast<int64_t>(value) * static_cast<int64_t>(rhs.value));
#endif
return *this;
}
template <typename U = T>
typename enable_if<is_same<typename Modular<U>::Type, long long>::value, Modular>::type& operator*=(const Modular& rhs) {
long long q = static_cast<long long>(static_cast<long double>(value) * rhs.value / mod());
value = normalize(value * rhs.value - q * mod());
return *this;
}
template <typename U = T>
typename enable_if<!is_integral<typename Modular<U>::Type>::value, Modular>::type& operator*=(const Modular& rhs) {
value = normalize(value * rhs.value);
return *this;
}
Modular& operator/=(const Modular& other) { return *this *= Modular(inverse(other.value, mod())); }
friend const Type& abs(const Modular& x) { return x.value; }
template <typename U>
friend bool operator==(const Modular<U>& lhs, const Modular<U>& rhs);
template <typename U>
friend bool operator<(const Modular<U>& lhs, const Modular<U>& rhs);
template <typename V, typename U>
friend V& operator>>(V& stream, Modular<U>& number);
private:
Type value;
};
template <typename T> bool operator==(const Modular<T>& lhs, const Modular<T>& rhs) { return lhs.value == rhs.value; }
template <typename T, typename U> bool operator==(const Modular<T>& lhs, U rhs) { return lhs == Modular<T>(rhs); }
template <typename T, typename U> bool operator==(U lhs, const Modular<T>& rhs) { return Modular<T>(lhs) == rhs; }
template <typename T> bool operator!=(const Modular<T>& lhs, const Modular<T>& rhs) { return !(lhs == rhs); }
template <typename T, typename U> bool operator!=(const Modular<T>& lhs, U rhs) { return !(lhs == rhs); }
template <typename T, typename U> bool operator!=(U lhs, const Modular<T>& rhs) { return !(lhs == rhs); }
template <typename T> bool operator<(const Modular<T>& lhs, const Modular<T>& rhs) { return lhs.value < rhs.value; }
template <typename T> Modular<T> operator+(const Modular<T>& lhs, const Modular<T>& rhs) { return Modular<T>(lhs) += rhs; }
template <typename T, typename U> Modular<T> operator+(const Modular<T>& lhs, U rhs) { return Modular<T>(lhs) += rhs; }
template <typename T, typename U> Modular<T> operator+(U lhs, const Modular<T>& rhs) { return Modular<T>(lhs) += rhs; }
template <typename T> Modular<T> operator-(const Modular<T>& lhs, const Modular<T>& rhs) { return Modular<T>(lhs) -= rhs; }
template <typename T, typename U> Modular<T> operator-(const Modular<T>& lhs, U rhs) { return Modular<T>(lhs) -= rhs; }
template <typename T, typename U> Modular<T> operator-(U lhs, const Modular<T>& rhs) { return Modular<T>(lhs) -= rhs; }
template <typename T> Modular<T> operator*(const Modular<T>& lhs, const Modular<T>& rhs) { return Modular<T>(lhs) *= rhs; }
template <typename T, typename U> Modular<T> operator*(const Modular<T>& lhs, U rhs) { return Modular<T>(lhs) *= rhs; }
template <typename T, typename U> Modular<T> operator*(U lhs, const Modular<T>& rhs) { return Modular<T>(lhs) *= rhs; }
template <typename T> Modular<T> operator/(const Modular<T>& lhs, const Modular<T>& rhs) { return Modular<T>(lhs) /= rhs; }
template <typename T, typename U> Modular<T> operator/(const Modular<T>& lhs, U rhs) { return Modular<T>(lhs) /= rhs; }
template <typename T, typename U> Modular<T> operator/(U lhs, const Modular<T>& rhs) { return Modular<T>(lhs) /= rhs; }
template<typename T, typename U>
Modular<T> power(const Modular<T>& a, const U& b) {
assert(b >= 0);
Modular<T> x = a, res = 1;
U p = b;
while (p > 0) {
if (p & 1) res *= x;
x *= x;
p >>= 1;
}
return res;
}
template <typename T>
bool IsZero(const Modular<T>& number) {
return number() == 0;
}
template <typename T>
string to_string(const Modular<T>& number) {
return to_string(number());
}
// U == std::ostream? but done this way because of fastoutput
template <typename U, typename T>
U& operator<<(U& stream, const Modular<T>& number) {
return stream << number();
}
// U == std::istream? but done this way because of fastinput
template <typename U, typename T>
U& operator>>(U& stream, Modular<T>& number) {
typename common_type<typename Modular<T>::Type, long long>::type x;
stream >> x;
number.value = Modular<T>::normalize(x);
return stream;
}
/*
using ModType = int;
struct VarMod { static ModType value; };
ModType VarMod::value;
ModType& md = VarMod::value;
using Mint = Modular<VarMod>;
*/
constexpr int md = 998244353;
using Mint = Modular<std::integral_constant<decay<decltype(md)>::type, md>>;
/*vector<Mint> fact(1, 1);
vector<Mint> inv_fact(1, 1);
Mint C(int n, int k) {
if (k < 0 || k > n) {
return 0;
}
while ((int) fact.size() < n + 1) {
fact.push_back(fact.back() * (int) fact.size());
inv_fact.push_back(1 / fact.back());
}
return fact[n] * inv_fact[k] * inv_fact[n - k];
}*/
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n, m, k;
cin >> n >> m >> k;
vector<Mint> f(m, 1);
for (int it = 1; it < n; it++) {
vector<Mint> pref(m + 1);
for (int i = 0; i < m; i++) {
pref[i + 1] = pref[i] + f[i];
}
for (int i = 0; i < m; i++) {
int L = max(0, i - k + 1);
int R = min(m - 1, i + k - 1);
if (k == 0) {
f[i] = pref[m];
} else {
f[i] = pref[m] - (pref[R + 1] - pref[L]);
}
}
}
cout << accumulate(f.begin(), f.end(), Mint(0)) << '\n';
return 0;
}
F - Operations on a Matrix
- https://atcoder.jp/contests/abc253/tasks/abc253_f
参考题解（膜拜大神）
#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif
template <typename T>
class fenwick {
public:
vector<T> fenw;
int n;
fenwick(int _n) : n(_n) {
fenw.resize(n);
}
void modify(int x, T v) {
while (x < n) {
fenw[x] += v;
x |= (x + 1);
}
}
T get(int x) {
T v{};
while (x >= 0) {
v += fenw[x];
x = (x & (x + 1)) - 1;
}
return v;
}
};
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n, m, q;
cin >> n >> m >> q;
vector<array<int, 4>> qs(q);
vector<int> when(n, 0);
vector<int> what(n, 0);
vector<long long> res(q, -1);
vector<vector<pair<int, int>>> ask(q);
fenwick<long long> fenw(m);
for (int i = 0; i < q; i++) {
auto& p = qs[i];
cin >> p[0] >> p[1] >> p[2];
if (p[0] == 1) {
cin >> p[3];
--p[1]; --p[2];
fenw.modify(p[1], p[3]);
fenw.modify(p[2] + 1, -p[3]);
} else {
if (p[0] == 2) {
--p[1];
when[p[1]] = i;
what[p[1]] = p[2];
} else {
--p[1]; --p[2];
res[i] = what[p[1]] + fenw.get(p[2]);
ask[when[p[1]]].emplace_back(p[2], i);
}
}
}
fenwick<long long> fenw2(m);
for (int i = 0; i < q; i++) {
for (auto& p : ask[i]) {
res[p.second] -= fenw2.get(p.first);
}
auto& p = qs[i];
if (p[0] == 1) {
fenw2.modify(p[1], p[3]);
fenw2.modify(p[2] + 1, -p[3]);
}
}
for (int i = 0; i < q; i++) {
if (qs[i][0] == 3) {
cout << res[i] << '\n';
}
}
return 0;
}
参考题解（官方提供）
#include <bits/stdc++.h>
#include <atcoder/fenwicktree>
using namespace std;
int main() {
int n, m, q;
cin >> n >> m >> q;
vector<int> t(q), a(q), b(q), c(q);
vector<vector<int>> subt(q);
vector latest(n, pair(-1, 0));
vector<long long> ans;
for (int i = 0; i < q; ++i) {
cin >> t[i];
if (t[i] == 1) {
cin >> a[i] >> b[i] >> c[i];
a[i] -= 1;
} else if (t[i] == 2) {
cin >> a[i] >> b[i];
a[i] -= 1;
latest[a[i]] = pair(i, b[i]);
} else {
cin >> a[i] >> b[i];
a[i] -= 1;
const auto& [j, x] = latest[a[i]];
const int id = ans.size();
ans.emplace_back(x);
c[i] = id;
if (j >= 0) {
subt[j].push_back(i);
}
}
}
atcoder::fenwick_tree<long long> fen(m + 1);
for (int i = 0; i < q; ++i) {
if (t[i] == 1) {
fen.add(a[i], c[i]);
fen.add(b[i], -c[i]);
} else if (t[i] == 2) {
for (const int j : subt[i]) {
ans[c[j]] -= fen.sum(0, b[j]);
}
} else {
ans[c[i]] += fen.sum(0, b[i]);
}
}
for (const long long x : ans) {
cout << x << '\n';
}
return 0;
}本文参与 腾讯云自媒体同步曝光计划，分享自微信公众号。
原始发表：2022-05-29，如有侵权请联系 cloudcommunity@tencent.com 删除
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