結果

問題 No.2360 Path to Integer
ユーザー tnakao0123
提出日時 2024-06-26 16:16:33
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 107 ms / 2,500 ms
コード長 3,278 bytes
コンパイル時間 1,065 ms
コンパイル使用メモリ 116,740 KB
最終ジャッジ日時 2025-02-22 00:40:01
ジャッジサーバーID
（参考情報） 		judge2 / judge2
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ファイルパターン 結果
sample AC * 2
other AC * 15
権限があれば一括ダウンロードができます
コンパイルメッセージ
main.cpp: In function ‘int main()’:
main.cpp:93:8: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   93 |   scanf("%d", &n);
      |   ~~~~~^~~~~~~~~~
main.cpp:97:10: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   97 |     scanf("%s", s);
      |     ~~~~~^~~~~~~~~
main.cpp:107:10: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
  107 |     scanf("%d%d", &u, &v);
      |     ~~~~~^~~~~~~~~~~~~~~~

ソースコード

diff #
プレゼンテーションモードにする

/* -*- coding: utf-8 -*-
 *
 * 2360.cc:  No.2360 Path to Integer - yukicoder
 */
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<deque>
#include<algorithm>
#include<numeric>
#include<utility>
#include<complex>
#include<functional>
 
using namespace std;
/* constant */
const int MAX_N = 100000;
const int MAX_L = 10;
const int MOD = 998244353;
const int INF = 1 << 30;
const long long LINF = 1LL << 60;
/* typedef */
using ll = long long;
using vi = vector<int>;
using qi = queue<int>;
using pii = pair<int,int>;
template<const int MOD>
struct MI {
  int v;
  MI(): v() {}
  MI(int _v): v(_v % MOD) { if (v < 0) v += MOD; }
  MI(long long _v): v(_v % MOD) { if (v < 0) v += MOD; }
  explicit operator int() const { return v; }
  
  MI operator+(const MI m) const { return MI(v + m.v); }
  MI operator-(const MI m) const { return MI(v + MOD - m.v); }
  MI operator*(const MI m) const { return MI((long long)v * m.v); }
  MI &operator+=(const MI m) { return (*this = *this + m); }
  MI &operator-=(const MI m) { return (*this = *this - m); }
  MI &operator*=(const MI m) { return (*this = *this * m); }
  bool operator==(const MI m) const { return v == m.v; }
  bool operator!=(const MI m) const { return v != m.v; }
  MI pow(int n) const {  // a^n % MOD
    MI pm = 1, a = *this;
    while (n > 0) {
      if (n & 1) pm *= a;
      a *= a;
      n >>= 1;
    }
    return pm;
  }
  MI inv() const { return pow(MOD - 2); }
  MI operator/(const MI m) const { return *this * m.inv(); }
  MI &operator/=(const MI m) { return (*this = *this / m); }
};
using mi = MI<MOD>;
/* global variables */
mi as[MAX_N], ts[MAX_L + 1];
int als[MAX_N];
vi nbrs[MAX_N];
int ps[MAX_N], cis[MAX_N], ss[MAX_N];
mi dp0[MAX_N], dp1[MAX_N];
/* subroutines */
/* main */
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    char s[MAX_L + 4];
    scanf("%s", s);
    int l = 0;
    while (s[l]) as[i] = as[i] * 10 + (s[l++] - '0');
    als[i] = l;
    //printf(" %d,%d", (int)as[i], als[i]);
  }
  //putchar('\n');
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    u--, v--;
    nbrs[u].push_back(v);
    nbrs[v].push_back(u);
  }
  ts[0] = 1;
  for (int i = 1; i <= MAX_L; i++) ts[i] = ts[i - 1] * 10;
  ps[0] = -1;
  for (int u = 0; u >= 0;) {
    auto &nbru = nbrs[u];
    int up = ps[u];
    if (cis[u] < nbru.size()) {
      int v = nbru[cis[u]++];
      if (v != up) {
    ps[v] = u;
    u = v;
      }
    }
    else {
      ss[u]++;
      dp0[u] += mi(as[u]) * ss[u];
      if (up >= 0) {
    ss[up] += ss[u];
    dp0[up] += dp0[u] * ts[als[up]];
      }
      u = up;
    }
  }
  fill(cis, cis + n, 0);
  dp1[0] = 0;
  for (int u = 0; u >= 0;) {
    auto &nbru = nbrs[u];
    int up = ps[u];
    if (cis[u] < nbru.size()) {
      int v = nbru[cis[u]++];
      if (v != up) {
    dp1[v] =
      (dp1[u] + dp0[u] - (dp0[v] * ts[als[u]] + mi(as[u]) * ss[v]))
       * ts[als[v]] +
      mi(as[v]) * (n - ss[v]);
    u = v;
      }
    }
    else {
      u = up;
    }
  }
  mi sum = 0;
  for (int u = 0; u < n; u++)
    sum += dp0[u] + dp1[u];
  printf("%d\n", (int)sum);
  
  return 0;
}
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