Submission #59340

#TimeUsernameProblemLanguageResultExecution timeMemory
59340BenqRelativnost (COCI15_relativnost)C++11
140 / 140
1408 ms24276 KiB
#include <bits/stdc++.h> #include <ext/pb_ds/tree_policy.hpp> #include <ext/pb_ds/assoc_container.hpp> using namespace std; using namespace __gnu_pbds; typedef long long ll; typedef long double ld; typedef complex<ld> cd; typedef pair<int, int> pi; typedef pair<ll,ll> pl; typedef pair<ld,ld> pd; typedef vector<int> vi; typedef vector<ld> vd; typedef vector<ll> vl; typedef vector<pi> vpi; typedef vector<pl> vpl; typedef vector<cd> vcd; template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>; #define FOR(i, a, b) for (int i=a; i<(b); i++) #define F0R(i, a) for (int i=0; i<(a); i++) #define FORd(i,a,b) for (int i = (b)-1; i >= a; i--) #define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--) #define sz(x) (int)(x).size() #define mp make_pair #define pb push_back #define f first #define s second #define lb lower_bound #define ub upper_bound #define all(x) x.begin(), x.end() const int MOD = 10007; const ll INF = 1e18; const int MX = 100001; int N,C; template<int SZ> struct Seg { short tot[2*SZ]; array<short,20> num[2*SZ]; Seg() { FOR(i,SZ,2*SZ) { tot[i] = 1; num[i][0] = 1; } } void comb(array<short,20>& a, array<short,20>& b, array<short,20>& c) { F0R(i,20) a[i] = 0; F0R(i,20) F0R(j,20-i) a[i+j] = (a[i+j]+b[i]*c[j])%MOD; } void pull(int x) { tot[x] = tot[2*x]*tot[2*x+1]%MOD; comb(num[x],num[2*x],num[2*x+1]); } void build() { FORd(i,1,SZ) pull(i); } void upd(int ind, pi x) { ind ^= 1<<17; num[ind][1] = x.f; num[ind][0] = x.s; tot[ind] = (num[ind][1]+num[ind][0])%MOD; for (ind /= 2; ind; ind /= 2) pull(ind); } }; Seg<1<<17> S; int main() { ios_base::sync_with_stdio(0); cin.tie(0); cin >> N >> C; F0R(i,N) { int x; cin >> x; S.num[i^(1<<17)][1] = x%MOD; } F0R(i,N) { int x; cin >> x; S.num[i^(1<<17)][0] = x%MOD; S.tot[i^(1<<17)] = (S.num[i^(1<<17)][0]+S.num[i^(1<<17)][1])%MOD; } S.build(); int Q; cin >> Q; F0R(i,Q) { int p,a,b; cin >> p >> a >> b; S.upd(p-1,{a%MOD,b%MOD}); short ans = S.tot[1]; // cout << "AH " << ans << "\n"; F0R(i,C) ans = (ans+MOD-S.num[1][i])%MOD; cout << ans << "\n"; } } /* Look for: * the exact constraints (multiple sets are too slow for n=10^6 :( ) * special cases (n=1?) * overflow (ll vs int?) * array bounds */
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