제출 #966634

#제출 시각아이디문제언어결과실행 시간메모리
966634angellaMisspelling (JOI22_misspelling)C++17
100 / 100
392 ms324396 KiB
/* IN THE NAME OF GOD */ #include <bits/stdc++.h> // #pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt") // #pragma GCC optimize("O3") // #pragma GCC optimize("unroll-loops") using namespace std; typedef long long ll; typedef pair<ll, ll> pll; typedef pair<int, int> pii; typedef long double ld; #define F first #define S second #define Mp make_pair #define pb push_back #define pf push_front #define size(x) ((ll)x.size()) #define all(x) (x).begin(),(x).end() #define kill(x) cout << x << '\n', exit(0); #define fuck(x) cout << "(" << #x << " , " << x << ")" << endl #define endl '\n' const int N = 5e5+23, lg = 18; ll Mod = 1e9+7; //998244353; inline ll MOD(ll a, ll mod=Mod) {a%=mod; (a<0)&&(a+=mod); return a;} inline ll poww(ll a, ll b, ll mod=Mod) { ll ans = 1; a=MOD(a, mod); while (b) { if (b & 1) ans = MOD(ans*a, mod); b >>= 1; a = MOD(a*a, mod); } return ans; } ll n, m, a[N], rmn[N], rmx[N], dp[27][N], pr[27][N], pr2[27][N]; int main () { ios_base::sync_with_stdio(false), cin.tie(0); cin>>n>>m; for(ll v,u,i=1; i<=m; i++) { cin>>v>>u; if(v<u) rmx[v] = max(rmx[v], u); else rmn[u] = max(rmn[u], v); } vector<int> smn={1}, smx={1}; for(int i=1; i<=26; i++) dp[i][1]=pr2[i][1]=1, pr[i][1]=i; for(int i=2; i<=n; i++) { while(size(smx)>0&&rmx[smx.back()]<i) smx.pop_back(); while(size(smn)>0&&rmn[smn.back()]<i) smn.pop_back(); int mx=(size(smx)==0?0:smx.back()); int mn=(size(smn)==0?0:smn.back()); if(mn >= mx) { for(int j=1; j<=26; j++) { dp[j][i] = MOD(pr[26][i-1]-pr[26][mn]-(pr2[j][i-1]-pr2[j][mn])+ pr[26][mn]-pr[26][mx]-(pr[j][mn]-pr[j][mx])); } } else { for(int j=1; j<=26; j++) { dp[j][i] = MOD(pr[26][i-1]-pr[26][mx]-(pr2[j][i-1]-pr2[j][mx])+ (pr[j-1][mx]-pr[j-1][mn])); } } for(int j=1; j<=26; j++) { pr2[j][i] = MOD(pr2[j][i-1] + dp[j][i]); pr[j][i] = MOD(pr2[j][i] + pr[j-1][i]); } while(size(smx)>0&&rmx[smx.back()]<=rmx[i]) smx.pop_back(); smx.pb(i); while(size(smn)>0&&rmn[smn.back()]<=rmn[i]) smn.pop_back(); smn.pb(i); } cout<<pr[26][n]<<endl; return 0; }
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