Submission #331020

#	Time	Username	Problem	Language	Result	Execution time	Memory
331020		leinad2	수열 (APIO14_sequence)	C++14	49 / 100	174 ms	131076 KiB

sequence

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
struct line
{
    ll a, b, index;
};
vector<line>V;
double cross(const line &p, const line &q)
{
    return (double)(q.b-p.b)/(p.a-q.a);
}
void push(line x)
{
    if(V.size()&&V.back().a==x.a)
    {
        if(V.back().b<x.b)V.back().b=x.b;
        return;
    }
    while(V.size()>=2)
    {
        if(cross(V.back(), V[V.size()-2])>cross(V.back(), x))V.pop_back();
        else break;
    }
    V.push_back(x);
}
pair<ll, ll>query(ll x)
{
    ll a=0, b=V.size(), m;
    while(a+1<b)
    {
        m=(a+b)/2;
        if(cross(V[m-1], V[m])<=x)a=m;
        else b=m;
    }
    return {V[a].a*x+V[a].b, V[a].index};
}
ll n, i, j, k, t, a, b, c, A[100010], dp[100010][2], backtrack[100010][210];
int main()
{
    for(scanf("%lld %lld", &n, &k);i++<n;)
    {
        scanf("%lld", &A[i]);
        A[i]+=A[i-1];
    }
    for(j=0;j++<k;)
    {
        for(i=0;i++<n;)
        {
            push({A[i], dp[i][0]-A[i]*A[i], i});
            pair<ll, ll>p=query(A[i]);
            dp[i][1]=p.first;
            backtrack[i][j]=p.second;
            dp[i][0]=dp[i][1];
        }
        V.clear();
    }
    printf("%lld\n", dp[n][1]);
    i=n;
    j=k;
    vector<ll>ans;
    while(j>0)
    {
        i=backtrack[i][j];
        j--;
        ans.push_back(i);
    }
    while(ans.size())
    {
        printf("%lld ", ans.back());
        ans.pop_back();
    }
}

Compilation message (stderr)

sequence.cpp: In function 'int main()':
sequence.cpp:41:14: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
   41 |     for(scanf("%lld %lld", &n, &k);i++<n;)
      |         ~~~~~^~~~~~~~~~~~~~~~~~~~~
sequence.cpp:43:14: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
   43 |         scanf("%lld", &A[i]);
      |         ~~~~~^~~~~~~~~~~~~~~

• Subtask 1: 0 / 11
• Subtask 2: 11 / 11
• Subtask 3: 0 / 11
• Subtask 4: 17 / 17
• Subtask 5: 21 / 21
• Subtask 6: 0 / 29