# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
365057 | qwerasdfzxcl | Split the sequence (APIO14_sequence) | C++14 | 1004 ms | 84220 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int a[100100];
ll s[100100], dp[100100], tmp[100100];
vector<int> fx;
long double ran[100100];
int n, k;
int parent[100100][202];
pair<long double, bool> po(int x1, int x2){
if (s[x1]==s[x2]) return make_pair(0, 0);
pair<long double, bool> ret=make_pair(0, 1);
ll tmp1=s[x2]*s[x2]-s[x1]*s[x1]+tmp[x2]-tmp[x1];
ll tmp2=s[x2]*2-s[x1]*2;
ret.first=(long double)tmp1/tmp2;
return ret;
}
void cht(int t){
int cur=0;
dp[0]=1e18;
for (int i=1;i<n;i++){
if (i<t){
dp[i]=1e18;
continue;
}
if (i==t){
fx.push_back(i-1);
dp[i]=s[i-1]*(-2)*s[i]+tmp[i-1]+(s[i-1]*s[i-1])+(s[i]*s[i]);
parent[i][t]=i-1;
continue;
}
pair<long double, bool> tmpp=po(fx.back(), i-1);
while(tmpp.second && tmpp.first<ran[(int)fx.size()-1]){
ran[(int)fx.size()-1]=1e20;
fx.pop_back();
tmpp=po(fx.back(), i-1);
}
if (tmpp.second){
ran[(int)fx.size()]=tmpp.first;
fx.push_back(i-1);
}
if (cur>=((int)fx.size()-1)){
cur=(int)fx.size()-1;
}
for (;cur<(int)fx.size()-1;cur++){
if ((long double)s[i]>ran[cur+1]) continue;
break;
}
for (;cur>0;cur--){
if ((long double)s[i]<ran[cur]) continue;
break;
}
assert(ran[cur]<=(long double)s[i] && (long double)s[i]<=ran[cur+1]);
//printf("%d %d\n", t, i);
//for (int j=0;j<=(int)fx.size();j++) printf("%Lf ", ran[j]);
//printf("\n");
dp[i]=s[fx[cur]]*(-2)*s[i]+tmp[fx[cur]]+(s[fx[cur]]*s[fx[cur]])+(s[i]*s[i]);
parent[i][t]=fx[cur];
}
for (int i=0;i<n;i++){
//printf("%lld ", dp[i]);
tmp[i]=dp[i];
}
//printf("\n");
}
int main(){
scanf("%d %d", &n, &k);
for (int i=0;i<n;i++) scanf("%d", a+i);
s[0]=a[0];
for (int i=1;i<n;i++) s[i]=s[i-1]+a[i];
for (int i=0;i<n;i++){
tmp[i]=s[i]*s[i];
}
for (int i=0;i<n;i++) parent[i][0]=-1;
for (int i=1;i<=k;i++){
fx.clear();
ran[0]=-1e20;
fill(ran+1, ran+n+3, 1e20);
cht(i);
}
printf("%lld\n", (s[n-1]*s[n-1]-dp[n-1])/2);
//printf("%d\n", parent[n-1][k]);
int idx=k-1;
for (int i=parent[n-1][k];i!=-1;i=parent[i][idx--]){
printf("%d ", i+1);
}
return 0;
}
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