| # | Time | Username | Problem | Language | Result | Execution time | Memory |
|---|---|---|---|---|---|---|---|
| 931525 | pan | Hop (COCI21_hop) | C++17 | 0 ms | 0 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>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#include "bits_stdc++.h"
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define input(x) scanf("%lld", &x);
#define input2(x, y) scanf("%lld%lld", &x, &y);
#define input3(x, y, z) scanf("%lld%lld%lld", &x, &y, &z);
#define input4(x, y, z, a) scanf("%lld%lld%lld%lld", &x, &y, &z, &a);
#define print(x, y) printf("%lld%c", x, y);
#define show(x) cerr << #x << " is " << x << endl;
#define show2(x,y) cerr << #x << " is " << x << " " << #y << " is " << y << endl;
#define show3(x,y,z) cerr << #x << " is " << x << " " << #y << " is " << y << " " << #z << " is " << z << endl;
#define discretize(x) sort(x.begin(), x.end()); x.erase(unique(x.begin(), x.end()), x.end());
using namespace std;
//using namespace __gnu_pbds;
//#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
//#define ordered_multiset tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update>
typedef long long ll;
typedef long double ld;
typedef pair<ld, ll> pd;
typedef pair<string, ll> psl;
typedef pair<ll, ll> pi;
typedef pair<ll, pi> pii;
int main()
{
ll n;
input(n);
ll arr[n];
pi dp[n][3];
ll ans[n][n];
for (ll i=0; i<n; ++i) dp[i][0] = mp(0,1), dp[i][1] = mp(0,2), dp[i][2] = mp(0,3);
for (ll i=0; i<n; ++i) input(arr[i]);
for (ll i=1; i<n; ++i)
{
for (ll j=0; j<i; ++j)
{
if (arr[i]%arr[j]!=0) {ans[j][i] = 1; continue; }// does not matter
sort(dp[j], dp[j]+3);
dp[j][0].f++;
ans[j][i] = dp[j][0].s;
for (ll k=0; k<3; ++k) dp[i][dp[j][k].s-1].f = max(dp[i][dp[j][k].s-1].f, dp[j][k].f);
}
}
for (ll i=1; i<n; ++i) {for (ll j=0; j<i; ++j) {print(ans[j][i], ' '); } printf("%c", '\n');}
return 0;
}