제출 #465307

#제출 시각아이디문제언어결과실행 시간메모리
465307arnob918수열 (APIO14_sequence)C++14
0 / 100
19 ms1988 KiB
#include <bits/stdc++.h> #include <stdio.h> #include <ext/pb_ds/assoc_container.hpp>// For pbds.Don't use tree as variable name. #include <ext/pb_ds/tree_policy.hpp> using namespace std; using namespace __gnu_pbds; #define pb push_back #define eb emplace_back #define mem(x,i) memset(x,i,sizeof(x)) #define ff first #define ss second #define all(x) x.begin(),x.end() #define Q int t; scanf("%d", &t); for(int q=1; q<=t; q++) typedef long long ll; typedef unsigned long long ull; typedef double ld;//%Lf typedef pair<ll, ll> pi; template <typename T> using orderedSet = tree<T, null_type, less<T>,rb_tree_tag, tree_order_statistics_node_update>; //order_of_key(k) - number of element strictly less than k. //find_by_order(k) - k'th element in set.(0 indexed)(iterator) /* Debug Tools */ #define error(args...) \ { \ string _s = #args; replace(_s.begin(), _s.end(), ',', ' ');\ stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args);\ } void err(istream_iterator<string> it) {} template<typename T, typename... Args> void err(istream_iterator<string> it, T a, Args... args) { cerr<< *it << " = " << a <<",\n"[++it==istream_iterator<string>()]; err(it, args...); } const int MOD = 1e9+7 ; //For big mod template<typename T>inline T gcd(T a, T b){T c;while (b){c = b;b = a % b;a = c;}return a;} // better than __gcd ll powmod(ll a,ll b){ll res=1;a%=MOD;if(b<0) return 0;for(; b; b>>=1){if(b&1)res=res*a%MOD;a=a*a%MOD;}return res;} //const int xx[] = {+1, -1, +0, +0};//, +1, +1, -1, -1};// exclude last four when side adjacent const int yy[] = {+0, +0, +1, -1};//, +1, -1, +1, -1}; const int INF = 0x3f3f3f3f;// useful for memset const ll LL_INF = 0x3f3f3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-9; const int mxn = 1e5+5; //CHECK here for every problem const int mod = 1e9+7; int n, k; vector<pi> v; ll a[mxn]; ll z[mxn]; pi mid(int b, int e){ int lo = b; int hi = e; int pos = lo; while(hi >= lo){ int m1 = lo+(hi-lo)/3; int m2 = hi-(hi-lo)/3; ll c1 = abs((a[m1]-(b==0? 0: a[b-1])) - (a[e]-a[m1])); ll c2 = abs((a[m2]-(b==0? 0: a[b-1])) - (a[e]-a[m2])); if(c2 > c1){ pos = m1; hi = m2-1; } else{ pos = m2; lo = m1+1; } } pi ret; ret.ff = (a[pos]-(b==0? 0: a[b-1])) * (a[e]-a[pos]); ret.ss = pos+1; return ret; } void call(int b=0, int e=n-1){ if(b==e) return; //error(b, e); pi val = mid(b, e); //error(val.ff, val.ss); v.pb(val); call(b, val.ss-1); call(val.ss, e); } int main() { cin >> n >> k; for(int i=0; i<n;i++){ cin >> a[i]; } for(int i=1; i<n; i++) a[i] += a[i-1]; int ans = 0; vector<int> aa; for(int msk = 0; msk<(1<<n); msk++){ if(__builtin_popcount(msk) != k)continue; int la = -1; int cnt = 0; vector<int> xx; for(int i=0; i<n; i++){ if(msk&(1<<i)){ if(la == -1){ cnt += a[i]*(a[n-1]-a[i]); la = i; } else{ cnt += (a[i]-a[la])*(a[n-1]-a[i]); la = i; } xx.pb(i+1); } } if(cnt > ans){ ans = cnt; aa = xx; } } cout << ans << '\n'; for(int i=0; i<k; i++){ printf("%d ", aa[i]); } printf("\n"); }
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