제출 #106590

#제출 시각아이디문제언어결과실행 시간메모리
106590ekrem수열 (APIO14_sequence)C++98
71 / 100
139 ms132096 KiB
#include <bits/stdc++.h> #define st first #define nd second #define mp make_pair #define pb push_back #define mod 1000000007 #define N 100005 using namespace std; typedef long long ll; typedef pair < int , ll > ii; ll dp[N][205]; int git[N][205], pre[N], a[N], n, k, bas, son; pair < int , ii > q[N]; inline double kesisim(ii a, ii b){return 1.0 * ( b.nd - a.nd ) / (1.0 * (a.st - b.st)) ;} void ekle(ll i, ii d){ if(son - bas + 1 >= 1 and d.st == q[son].nd.st){ if(d.nd < q[son].nd.nd) return; else son--; } while(son - bas + 1 >= 2){ ii d1 = q[son].nd; ii d2 = q[son - 1].nd; if(kesisim(d, d2) <= kesisim(d1, d2)) son--; else break; } q[++son] = mp(i, d); } ll bul(ll x){ while(son - bas + 1 >= 2){ ii d1 = q[bas].nd; ii d2 = q[bas + 1].nd; if(kesisim(d1, d2) <= x) bas++; else break; } return q[bas].st; } int main() { // freopen("in.txt", "r", stdin); // freopen("out.txt", "w", stdout); scanf("%d %d",&n ,&k); for(int i = 1; i <= n; i++){ scanf("%d",a + i); pre[i] = pre[i - 1] + a[i]; } for(int x = 1; x <= k; x++){ bas = 1, son = 0; ekle(0, mp(0, 0)); for(int i = 1; i <= n; i++){ ll j = bul(pre[i]); dp[i][x] = dp[j][x - 1] + 1ll*pre[j]*pre[i] - 1ll*pre[j]*pre[j]; git[i][x] = j; ekle(i, mp(pre[i], dp[i][x - 1] - 1ll*pre[i]*pre[i]) ); } } printf("%lld\n", dp[n][k]); son = n; bas = k; // if(dp[n][k] == 0){ // for(int i = 1; i <= k; i++) // printf("%d ", i); // return 0; // } for(ll i = 1; i <= k; i++){ printf("%d ", git[son][bas]); son = git[son][bas]; bas--; } return 0; }

컴파일 시 표준 에러 (stderr) 메시지

sequence.cpp: In function 'int main()':
sequence.cpp:52:7: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
  scanf("%d %d",&n ,&k);
  ~~~~~^~~~~~~~~~~~~~~~
sequence.cpp:54:8: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
   scanf("%d",a + i);
   ~~~~~^~~~~~~~~~~~
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