제출 #126007

#제출 시각아이디문제언어결과실행 시간메모리
126007kjp4155수열 (APIO14_sequence)C++17
100 / 100
563 ms87952 KiB
#include <bits/stdc++.h> using namespace std; #define VA_NUM_ARGS(...) VA_NUM_ARGS_IMPL_((0,__VA_ARGS__, 6,5,4,3,2,1)) #define VA_NUM_ARGS_IMPL_(tuple) VA_NUM_ARGS_IMPL tuple #define VA_NUM_ARGS_IMPL(_0,_1,_2,_3,_4,_5,_6,N,...) N #define macro_dispatcher(macro, ...) macro_dispatcher_(macro, VA_NUM_ARGS(__VA_ARGS__)) #define macro_dispatcher_(macro, nargs) macro_dispatcher__(macro, nargs) #define macro_dispatcher__(macro, nargs) macro_dispatcher___(macro, nargs) #define macro_dispatcher___(macro, nargs) macro ## nargs #define Debug1(a) cout<<#a<<"="<<(a)<<"\n" #define Debug2(a,b) cout<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<"\n" #define Debug3(a,b,c) cout<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<"\n" #define Debug4(a,b,c,d) cout<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<", "<<#d<<"="<<(d)<<"\n" #define Debug5(a,b,c,d,e) cout<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<", "<<#d<<"="<<(d)<<", "<<#e<<"="<<(e)<<"\n" #define Debug6(a,b,c,d,e,f) cout<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<", "<<#d<<"="<<(d)<<", "<<#e<<"="<<(e)<<", "<<#f<<"="<<(f)<<"\n" #define Debug(...) macro_dispatcher(Debug, __VA_ARGS__)(__VA_ARGS__) #define DA(a,s,n) cout<<#a<<"=["; printarray(a,s,n); cout<<"]\n" #define TT1 template<class T> #define TT1T2 template<class T1, class T2> #define TT1T2T3 template<class T1, class T2, class T3> template<class T, size_t N> ostream& operator << (ostream& os, const array<T, N>& v); TT1T2 ostream& operator << (ostream& os, const pair<T1, T2>& p){ return os <<"("<<p.first<<", "<< p.second<<")"; } TT1 ostream& operator << (ostream& os, const vector<T>& v){ bool f=1;os<<"[";for(auto& i : v) { if (!f)os << ", ";os<<i;f=0;}return os << "]"; } template<class T, size_t N> ostream& operator << (ostream& os, const array<T, N>& v) { bool f=1;os<<"[";for(auto& i : v) { if (!f)os << ", ";os<<i;f=0;}return os << "]"; } TT1T2 ostream& operator << (ostream& os, const set<T1, T2>&v){ bool f=1;os<<"[";for(auto& i : v) { if (!f)os << ", ";os<<i;f=0;}return os << "]"; } TT1T2 ostream& operator << (ostream& os, const multiset<T1,T2>&v){bool f=1;os<<"[";for(auto& i : v) { if (!f)os << ", ";os<<i;f=0;}return os << "]"; } TT1T2T3 ostream& operator << (ostream& os, const map<T1,T2,T3>& v){ bool f = 1; os << "["; for (auto& ii : v) { if (!f)os << ", "; os << "(" << ii.first << " -> " << ii.second << ") "; f = 0; }return os << "]"; } TT1T2 ostream& operator << (ostream& os, const multimap<T1, T2>& v){ bool f = 1; os << "["; for (auto& ii : v) { if (!f)os << ", "; os << "(" << ii.first << " -> " << ii.second << ") "; f = 0; }return os << "]"; } TT1T2 ostream& operator << (ostream& os, priority_queue<T1, T2> v) { bool f = 1; os << "["; while (!v.empty()) { auto x = v.top(); v.pop(); if (!f) os << ", "; f = 0; os << x; } return os << "]"; } TT1T2 void printarray(const T1& a, T2 l, T2 r){ for (T2 i = l; i<=r; i++) cout << a[i] << (i<r?" ":""); } void cio(){ ios::sync_with_stdio(false); cin.tie(NULL); cout << setprecision(15); } #define tt() printf("%.4f sec\n", (double) clock() / CLOCKS_PER_SEC ) typedef long long ll; typedef unsigned long long ull; typedef pair<int, int> pii; typedef pair<ll, ll> pll; #define Fi first #define Se second #define pb(x) push_back(x) #define sz(x) (int)x.size() #define rep(i, n) for(int i=0;i<n;i++) #define repp(i, n) for(int i=1;i<=n;i++) #define all(x) x.begin(), x.end() #define geti1(X) cin >> X #define geti2(X,Y) cin >> X >> Y #define geti3(X,Y,Z) cin >> X >> Y >> Z #define geti4(X,Y,Z,W) cin >> X >> Y >> Z >> W #define GET_MACRO(_1,_2,_3,_4,NAME,...) NAME #define geti(...) GET_MACRO(__VA_ARGS__, geti4, geti3, geti2, geti1) (__VA_ARGS__) #define endl '\n' typedef tuple<ll,ll,ll> t3; typedef tuple<ll,ll,ll,ll> t4; void die(){printf("-1\n"); exit(0);} int xx[8] = {1,-1,0,0,1,1,-1,-1}, yy[8] = {0,0,1,-1,1,-1,1,-1}; const ll mod = 1e9+9; int N,K; ll a[100010], S[100010]; ll dp[100010]; int prv[205][100010]; struct Line{ ll a,b,idx; }; Line dq[100010]; Line line[2][100010]; int l, r; inline ll f(Line a, ll x){ return a.a * x + a.b; } int main(){ scanf("%d%d",&N,&K); repp(i,N) scanf("%lld",&a[i]); repp(i,N) S[i] = S[i-1] + a[i]; for(int k=0;k<=K;k++){ l = r = 0; for(int i=k+1;i<=N;i++){ if( k == 0 ) dp[i] = 0; else{ // insert at back dq[r++] = line[k&1][i-1]; while( r-l >= 3 ){ int i1 = r-1, i2 = r-2, i3 = r-3; if( (dq[i3].b-dq[i1].b)*(dq[i2].a-dq[i3].a) <= (dq[i3].b-dq[i2].b)*(dq[i1].a-dq[i3].a) ){ r--; dq[r-1] = dq[r]; } else break; } // proceed at front while( r-l >= 2 ){ if( f(dq[l],S[i]) <= f(dq[l+1],S[i]) ) l++; else break; } dp[i] = f(dq[l], S[i]); prv[k][i] = dq[l].idx; } Line nl = {S[i], dp[i]-S[i]*S[i], i}; line[!(k&1)][i] = nl; } //for(int i=k+1;i<=N;i++) line[i] = nline[i]; } printf("%lld\n",dp[N]); vector<int> ansv; int idx = N; for(int k=K;k>=1;k--){ ansv.push_back(prv[k][idx]); idx = prv[k][idx]; if( !( 1 <= idx && idx <= N-1 ) ) break; } sort(all(ansv)); for(auto e : ansv) printf("%d ",e); printf("\n"); //tt(); }

컴파일 시 표준 에러 (stderr) 메시지

sequence.cpp: In function 'int main()':
sequence.cpp:131:5: warning: this 'for' clause does not guard... [-Wmisleading-indentation]
     for(auto e : ansv) printf("%d ",e); printf("\n");
     ^~~
sequence.cpp:131:41: note: ...this statement, but the latter is misleadingly indented as if it were guarded by the 'for'
     for(auto e : ansv) printf("%d ",e); printf("\n");
                                         ^~~~~~
sequence.cpp:87:10: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
     scanf("%d%d",&N,&K);
     ~~~~~^~~~~~~~~~~~~~
sequence.cpp:88:20: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
     repp(i,N) scanf("%lld",&a[i]);
               ~~~~~^~~~~~~~~~~~~~
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