結果

問題 No.619 CardShuffle
ユーザー sak
提出日時 2020-09-06 13:00:28
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 254 ms / 3,000 ms
コード長 3,227 bytes
コンパイル時間 2,414 ms
コンパイル使用メモリ 204,820 KB
最終ジャッジ日時 2025-01-14 07:42:23
ジャッジサーバーID
（参考情報） 		judge5 / judge4
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 1
other AC * 35
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> p_ll;
template<class T>
void debug(T itr1, T itr2) { auto now = itr1; while(now<itr2) { cout << *now << " "; now++; } cout << endl; }
#define repr(i,from,to) for (ll i=(ll)from; i<(ll)to; i++)
#define all(vec) vec.begin(), vec.end()
#define rep(i,N) repr(i,0,N)
#define per(i,N) for (int i=(int)N-1; i>=0; i--)
const ll MOD = pow(10,9)+7;
const ll LLINF = pow(2,61)-1;
const int INF = pow(2,30)-1;
vector<ll> fac;
void c_fac(int x=pow(10,6)+10) { fac.resize(x,true); rep(i,x) fac[i] = i ? (fac[i-1]*i)%MOD : 1; }
ll inv(ll a, ll m=MOD) { ll b = m, x = 1, y = 0; while (b!=0) { int d = a/b; a -= b*d; swap(a,b); x -= y*d; swap(x,y); } return (x+m)%m; }
ll nck(ll n, ll k) { return fac[n]*inv(fac[k]*fac[n-k]%MOD)%MOD; }
ll gcd(ll a, ll b) { if (a<b) swap(a,b); return b==0 ? a : gcd(b, a%b); }
ll lcm(ll a, ll b) { return a/gcd(a,b)*b; }
// ----------------------------------------------------------------------
// ----------------------------------------------------------------------
struct ope { 
  ll x, y, z; 
  ope operator+ (const ope &o) const { 
    ope result;
    if (z==-1) result = {x*o.x%MOD, o.y, o.z};
    else {
      if (o.z==-1) result = {x, y, z*o.x%MOD};
      else result = {x, (y+z*o.x+o.y)%MOD, o.z};
    }
    return result; 
  };
};
struct query { char t; ll x, y; };
struct SegTree {
  int size;
  vector<ope> pos;
  SegTree(ll N) { size = 1; while(size<N) size<<=1; pos.resize(2*size,{1,-1,-1}); }
  ope operator[](const ll &x) const { return pos[x+size]; }
  void build() { per(i,size) operate(i); }
  void set(ll x, const ope v) { pos[x+size] = v; }
  void update(ll x, const ope v) { set(x,v); x+=size; while (x>>=1) operate(x); }
  ope query(ll a, ll b) {
    ope L = {1,-1,-1}, R = {1,0,0};
    for (a+=size, b+=size; a<b; a>>=1, b>>=1) {
      if (a&1) { L = q(L,pos[a]); a++; }
      if (b&1) { b--; R = q(pos[b],R); }
    }
    return q(L,R);
  }
  void operate(ll i) { pos[i] = pos[i*2] + pos[i*2+1]; }
  ope q(ope x, ope y) { return x + y; }
};
// ----------------------------------------------------------------------
// ----------------------------------------------------------------------
int main() {
  ll N; cin >> N;
  ll C[N+1] = {}; C[0] = 1;
  rep(i,N) {
    if (i%2==0) cin >> C[i+1];
    else {
      char tc; cin >> tc;
      C[i+1] = tc=='+';
    }
  }
  // debug(C,C+(N+1));
  SegTree st((N+1)/2);
  rep(i,(N+1)/2) {
    if (C[i*2]==0) st.update(i, (ope){C[i*2+1], -1, -1});
    else st.update(i, (ope){1, 0, C[i*2+1]});
  }
  ll Q; cin >> Q;
  query q[Q]; rep(i,Q) cin >> q[i].t >> q[i].x >> q[i].y;
  rep(i,Q) {
    ll xp = q[i].x/2, yp = q[i].y/2;
    if (q[i].t=='!') {
      swap(C[q[i].x], C[q[i].y]);
      if (C[xp*2]==0) st.update(xp, (ope){C[xp*2+1], -1, -1});
      else st.update(xp, (ope){1, 0, C[xp*2+1]});
      if (C[yp*2]==0) st.update(yp, (ope){C[yp*2+1], -1, -1});
      else st.update(yp, (ope){1, 0, C[yp*2+1]});
    }
    else {
      ope o = (ope){0,0,C[xp*2+1]} + st.query(xp+1,yp+1);
      ll result = o.y;
      cout << result << endl;
      // cout << o.x << " " << o.y << " " << o.z << endl;
    }
  }
  // cout << result << endl;
  return 0;
}
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