結果
問題 | No.2229 Treasure Searching Rod (Hard) |
ユーザー |
|
提出日時 | 2023-02-24 22:55:27 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 58 ms / 2,000 ms |
コード長 | 4,314 bytes |
コンパイル時間 | 1,668 ms |
コンパイル使用メモリ | 168,820 KB |
実行使用メモリ | 6,944 KB |
最終ジャッジ日時 | 2024-09-13 05:57:43 |
合計ジャッジ時間 | 4,152 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge4 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 29 |
ソースコード
#include <bits/stdc++.h>#define all(v) v.begin(), v.end()#define rall(v) v.rbegin(), v.rend()#define rep(i,n) for(int i=0;i<(int)(n);i++)#define codefor int test;cin>>test;while(test--)#define INT(...) int __VA_ARGS__;in(__VA_ARGS__)#define LL(...) ll __VA_ARGS__;in(__VA_ARGS__)#define vector2d(type,name,h,...) vector<vector<type>>name(h,vector<type>(__VA_ARGS__))#define vector3d(type,name,h,w,...) vector<vector<vector<type>>>name(h,vector<vector<type>>(w,vector<type>(__VA_ARGS__)))using namespace std;using ll = long long;template<class T> using rpriority_queue = priority_queue<T, vector<T>, greater<T>>;template<class T> istream& operator>>(istream& is, vector<T>& vec) {for(T& x : vec)is >> x;return is;}template<class T> ostream& operator<<(ostream& os, const vector<T>& vec) {if(vec.empty())return os;os << vec[0];for(auto it = vec.begin(); ++it!= vec.end();)os << ' ' << *it;return os;}void in(){}template <class Head, class... Tail> void in(Head& head, Tail&... tail){cin >> head;in(tail...);}void out(){cout << '\n';}template<class T>void out(const T& a){cout << a << '\n';}template <class Head, class... Tail> void out(const Head& head,const Tail&... tail){cout << head << ' ';out(tail...);}const int INF = 1 << 30;const long long INF2 = 1ll << 60;template<class T> void chmax(T &a,const T b){if(b>a)a=b;}template<class T> void chmin(T &a,const T b){if(b<a)a=b;}template<const unsigned int MOD> struct prime_modint {using mint = prime_modint;unsigned int v;prime_modint() : v(0) {}prime_modint(unsigned int a) { a %= MOD; v = a; }prime_modint(int a) { a %= (int)(MOD); if(a < 0)a += MOD; v = a; }prime_modint(long long a) { a %= (int)(MOD); if(a < 0)a += MOD; v = a; }static constexpr int mod() { return MOD; }mint& operator++() {v++; if(v == MOD)v = 0; return *this;}mint& operator--() {if(v == 0)v = MOD; v--; return *this;}mint operator++(int) { mint result = *this; ++*this; return result; }mint operator--(int) { mint result = *this; --*this; return result; }mint& operator+=(const mint& rhs) { v += rhs.v; if(v >= MOD) v -= MOD; return *this; }mint& operator-=(const mint& rhs) { if(v < rhs.v) v += MOD; v -= rhs.v; return *this; }mint& operator*=(const mint& rhs) {v = (unsigned int)((unsigned long long)(v) * rhs.v % MOD);return *this;}mint& operator/=(const mint& rhs) { return *this = *this * rhs.inv(); }mint operator+() const { return *this; }mint operator-() const { return mint() - *this; }mint pow(long long n) const {assert(0 <= n);mint r = 1, x = *this;while (n) {if (n & 1) r *= x;x *= x;n >>= 1;}return r;}mint inv() const { assert(v); return pow(MOD - 2); }friend mint operator+(const mint& lhs, const mint& rhs) { return mint(lhs) += rhs; }friend mint operator-(const mint& lhs, const mint& rhs) { return mint(lhs) -= rhs; }friend mint operator*(const mint& lhs, const mint& rhs) { return mint(lhs) *= rhs; }friend mint operator/(const mint& lhs, const mint& rhs) { return mint(lhs) /= rhs; }friend bool operator==(const mint& lhs, const mint& rhs) { return (lhs.v == rhs.v); }friend bool operator!=(const mint& lhs, const mint& rhs) { return (lhs.v != rhs.v); }friend std::ostream& operator << (std::ostream &os, const mint& rhs) noexcept { return os << rhs.v; }};//using mint = prime_modint<1000000007>;using mint = prime_modint<998244353>;//等差数列の和を求める(初項,末項,項数)long long int arith_sum2(long long int a,long long int l,long long int n){return n * (a + l) / 2;}int main(){ios::sync_with_stdio(false);cin.tie(0);LL(h, w, k);ll y, x, v;mint ans;for(int i = 0; i < k; i++){cin >> y >> x >> v;mint d;d += mint(v) * y;if(x > 1){ll n = x - 1 + min(0ll, y - x);ll s = max(0ll, y - x) + 1;d += mint(v) * arith_sum2(s, y - 1, n);}if(x + 1 <= w){ll n = w - x + min(0ll, y + x - (w + 1));ll s = max(0ll, y + x - (w + 1)) + 1;d += mint(v) * arith_sum2(s, y - 1, n);}ans += d;}out(ans);}