제출 #1135030

#제출 시각아이디문제언어결과실행 시간메모리
1135030Lemser디지털 회로 (IOI22_circuit)C++20
4 / 100
327 ms10644 KiB
#include "circuit.h" #include <bits/stdc++.h> using namespace std; using ll = long long; using ull = unsigned long long; using lld = long double; using ii = pair<int,int>; using pll = pair<ll, ll>; using vi = vector<int>; using vll = vector<ll>; using vii = vector<ii>; using vpll = vector<pll>; using vlld = vector<lld>; // #define endl '\n' #define all(x) x.begin(),x.end() #define lsb(x) x&(-x) #define gcd(a,b) __gcd(a,b) #define sz(x) (int)x.size() #define mp make_pair #define pb push_back #define fi first #define se second #define fls cout.flush() #define fore(i,l,r) for(auto i=l;i<r;i++) #define fo(i,n) fore(i,0,n) #define forex(i,r,l) for(auto i=r; i>=l;i--) #define ffo(i,n) forex(i,n-1,0) bool cmin(ll &a, ll b) { if(b<a){a=b;return 1;}return 0; } bool cmax(ll &a, ll b) { if(b>a){a=b;return 1;}return 0; } void valid(ll in) { cout<<((in)?"Yes\n":"No\n"); } ll lcm(ll a, ll b) { return (a/gcd(a,b))*b; } ll gauss(ll n) { return (n*(n+1))/2; } const int N = 2e5 + 7; const int mod = 1e9 + 2022; vector<vll> graph; ll p[N], a[N], n, m, dp[N][2]; /* dp[u][0] = #acomodos de parametros de los del subarbol de u para que u sea 0 dp[u][1] = lo mismo para que u sea 1 */ void dfs (ll u) { if (u >= n) { dp[u][a[u]] = 1; dp[u][a[u]^1] = 0; return; } vll dpu(sz(graph[u])+1, 0); // dpu[i] = #formas de acomodar parametros para que u tenga i hijos // con 1's dpu[0] = 1; for (ll v: graph[u]) { dfs(v); // mezclar vll ndp(sz(graph[u])+1, 0); fo (i, sz(graph[u])+1) { (ndp[i] += dpu[i] * dp[v][0]) %= mod; if (i < sz(graph[u])) (ndp[i+1] += dpu[i] * dp[v][1]) %= mod; } dpu = ndp; } fore (i, 1, sz(graph[u])+1) { // poner de parametro i en u fore (j, i, sz(graph[u])+1) (dp[u][1] += dpu[j]) %= mod; fo (j, i) (dp[u][0] += dpu[j]) %= mod; } } void init(int N, int M, std::vector<int> P, std::vector<int> A) { n = N; m = M; graph = vector<vll>(N+M); fore (i, 1, N+M) graph[P[i]].pb(i); fo (i, N+M) p[i] = P[i]; fore (i, N, N+M) a[i] = A[i-N]; dfs(0); } int count_ways(int l, int r) { ll u = l; a[u] ^= 1; dp[u][a[u]] = 1; dp[u][a[u]^1] = 0; while (u > 0) { u = p[u]; ll dpu[3]{}; dpu[0] = 1; for (ll v: graph[u]) { (dpu[2] = dpu[2] * dp[v][0] + dpu[1] * dp[v][1]) %= mod; (dpu[1] = dpu[1] * dp[v][0] + dpu[0] * dp[v][1]) %= mod; (dpu[0] = dpu[0] * dp[v][0]) %= mod; } (dp[u][1] = dpu[1] + 2ll*dpu[2]) %= mod; (dp[u][0] = 2ll*dpu[0] + dpu[1]) %= mod; } return dp[0][1]; }
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