SRM650 div1 easy TaroFillingAStringDiv1

考えたこと

• 両端が同じ文字で間が偶数、両端が違う文字で間が奇数ならちゃんと埋めれば0にできて1パターンしかない
• これ以外の場合どうやっても+1は回避できない
• +1になるパターン数を求めて掛けていけばよさそう
• 実験したら(間の数+1)パターンになった
• MODを忘れずに掛けていったら通った

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<ll> VL;
typedef vector<VL> VVL;
typedef pair<int, int> PII;

#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(x) x.begin(), x.end()
#define IN(a, b, x) (a<=x&&x<b)
#define MP make_pair
#define PB push_back
const int INF = (1LL<<30);
const ll LLINF = (1LL<<60);
const double PI = 3.14159265359;
const double EPS = 1e-12;
const int MOD = 1000000007;
//#define int ll

template <typename T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template <typename T> T &chmax(T &a, const T &b) { return a = max(a, b); }

int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};

template<unsigned MOD>
class ModInt {
public:
  unsigned x;
  ModInt(): x(0) { }
  ModInt(signed y) : x(y >= 0 ? y % MOD : MOD - (-y) % MOD) {}
  unsigned get() const { return x; }

  // 逆数
  ModInt inv() const {
    ll a = 1, p = x, e = MOD-2;
    while(e > 0) {
      if(e%2 == 0) {p = (p*p) % MOD; e /= 2;}
      else {a = (a*p) % MOD; e--;}
    }
    a %= MOD;
    return ModInt(a);
  }
  // e乗
  ModInt pow(ll e) {
    ll a = 1, p = x;
    while(e > 0) {
      if(e%2 == 0) {p = (p*p) % MOD; e /= 2;}
      else {a = (a*p) % MOD; e--;}
    }
    a %= MOD;
    return ModInt(a);
  }
  // 2のx乗
  ModInt pow2() {
    ll a = 1, p = 2, e = x;
    while(e > 0) {
      if(e%2 == 0) {p = (p*p) % MOD; e /= 2;}
      else {a = (a*p) % MOD; e--;}
    }
    a %= MOD;
    return ModInt(a);
  }

  // Comparators
  bool operator <(ModInt b) { return x < b.x; }
  bool operator >(ModInt b) { return x > b.x; }
  bool operator<=(ModInt b) { return x <= b.x; }
  bool operator>=(ModInt b) { return x >= b.x; }
  bool operator!=(ModInt b) { return x != b.x; }
  bool operator==(ModInt b) { return x == b.x; }

  // increment, decrement
  ModInt operator++() { x++; return *this; }
  ModInt operator--() { x--; return *this; }

  // Basic Operations
  ModInt &operator+=(ModInt that) {
    x = ((ll)x+that.x)%MOD;
    return *this;
  }
  ModInt &operator-=(ModInt that) {
    x = ((((ll)x-that.x)%MOD)+MOD)%MOD;
    return *this;
  }
  ModInt &operator*=(ModInt that) {
    x = (ll)x * that.x % MOD;
    return *this;
  }
  // O(log(mod))かかるので注意
  ModInt &operator/=(ModInt that) {
    x = (ll)x * that.inv() % MOD;
    return *this;
  }
  ModInt &operator%=(ModInt that) {
    x = (ll)x % that.x;
    return *this;
  }
  ModInt operator+(ModInt that)const{return ModInt(*this) += that;}
  ModInt operator-(ModInt that)const{return ModInt(*this) -= that;}
  ModInt operator*(ModInt that)const{return ModInt(*this) *= that;}
  ModInt operator/(ModInt that)const{return ModInt(*this) /= that;}
  ModInt operator%(ModInt that)const{return ModInt(*this) %= that;}
};
typedef ModInt<1000000007> mint;
// Input/Output
ostream &operator<<(ostream& os, mint a) { return os << a.x; }
istream &operator>>(istream& is, mint &a) { return is >> a.x; }

class TaroFillingAStringDiv1 {
   public:
   int getNumber(int N, vector <int> pos, string val)
  {
    int n = pos.size();

    pair<int, char> p[55];
    REP(i, n) p[i] = {pos[i]-1, val[i]};
    sort(p, p+n);

    mint ret = 1;
    FOR(i, 1, n) {
      if((p[i].first - p[i-1].first - 1) % 2) {
        if(p[i-1].second != p[i].second) {
          ret *= (p[i].first - p[i-1].first);
        }
      } else {
        if(p[i-1].second == p[i].second) {
          ret *= (p[i].first - p[i-1].first);
        }
      }
    }

    return ret.x;
  }
};