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/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;
typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;
#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
const ll MOD = 9999999999999937LL;
const ll INF = 1e18;
const int MX = 100001;
const int B = 280;
int N, ans[MX];
ll cum[MX], sum[B];
vpl tmp[B];
unordered_set<ll> pos[B];
ll sub(ll a, ll b) { return (a-b+MOD)%MOD; }
ll ad(ll a, ll b) { return (a+b)%MOD; }
void rebuild() {
vpl al;
F0R(i,B) {
al.insert(al.end(),all(tmp[i]));
tmp[i].clear(); pos[i].clear();
pos[i].insert(0);
}
F0R(i,B) {
int l = (ll)sz(al)*i/B, r = (ll)sz(al)*(i+1)/B;
ll csum = 0;
FOR(j,l,r) {
tmp[i].pb(al[j]);
pos[i].insert(csum);
csum = ad(csum,al[j].f);
}
sum[i] = csum;
}
}
void insBucket(int num, ll a, pl b) {
int ind = 0;
while (a) a = sub(a,tmp[num][ind++].f);
tmp[num].insert(tmp[num].begin()+ind,b);
pos[num].clear();
ll csum = 0;
for (auto a: tmp[num]) {
pos[num].insert(csum);
// cout << "OH " << csum << "\n";
csum = ad(csum,a.f);
}
sum[num] = csum;
}
void ins(ll a, pl b) {
F0R(i,B) {
if (pos[i].count(a) || i == B-1) {
// cout << "HA " << a << " " << i << " " << b.f << " " << b.s << "\n";
insBucket(i,a,b);
return;
} else a = sub(a,sum[i]);
}
// cout << "OOPS " << a << "\n";
}
ll input() {
ll cur = 0;
string s; cin >> s;
for (char c: s) cur = (10*cur+(c-'0'))%MOD;
return cur;
}
int main() {
ios_base::sync_with_stdio(0); cin.tie(0);
F0R(i,B) pos[i].insert(0);
cin >> N;
FOR(i,1,N+1) {
cum[i] = input();
ll des = sub(cum[i],cum[i-1]);
ins(des-1,{des,i});
if (i % B == 0) rebuild();
}
int co = 0;
F0R(i,B) for (auto a: tmp[i]) ans[a.s] = ++co;
FOR(i,1,N+1) cout << ans[i] << " ";
}
/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/
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