# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
45546 | gnoor | 수열 (APIO14_sequence) | C++17 | 626 ms | 83860 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 <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
long long tbl[100100];
struct line {
long long m,c;
int id;
long long eval(long long x) {
return m*x+c;
}
};
bool bad(const line &a, const line &b, const line &c) {
//return a-c is left to a-b
return (b.c-a.c)*(a.m-c.m)>=(c.c-a.c)*(a.m-b.m);
}
long long dp[2][100100];
int par[210][100100];
int main () {
int n,k;
scanf("%d%d",&n,&k);
for (int i=1;i<=n;i++) {
scanf("%lld",&tbl[i]);
tbl[i]+=tbl[i-1];
}
vector<line> stk;
int cur,mn;
line tmp;
for (int j=1;j<=k;j++) {
stk.clear();
cur=j;
mn=0;
for (int i=j+1;i<=n;i++) {
while (cur<i) {
tmp.m=tbl[cur];
tmp.c=dp[(j-1)%2][cur]-tbl[cur]*tbl[cur];
tmp.id=cur;
while (stk.size()>=2&&bad(stk[stk.size()-2],stk.back(),tmp)) stk.pop_back();
stk.push_back(tmp);
cur++;
}
mn=min(mn,(int)stk.size()-1);
while (mn+1<(int)stk.size()&&stk[mn+1].eval(tbl[i])>=stk[mn].eval(tbl[i])) mn++;
dp[j%2][i]=stk[mn].eval(tbl[i]);
par[j][i]=stk[mn].id;
}
}
printf("%lld\n",dp[k%2][n]);
int nn=n;
for (int i=k;i>0;i--) {
printf("%d ",par[i][nn]);
nn=par[i][nn];
}
printf("\n");
return 0;
}
Compilation message (stderr)
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