#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/rope>
using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;
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 int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;
int n, k, a[51], dif[51], tmp[51];
vpi op;
void cut() {
int mn = MOD; FOR(i,1,n+1) mn = min(mn,a[i]);
FOR(i,1,n+1) a[i] -= mn;
}
void nor() {
while (1) {
int mx = 0; FOR(i,1,n+1) mx = max(mx,a[i]);
if (mx < k) return;
FOR(i,1,n+1) if (a[i] == 0) {
op.pb({1,i});
a[i] += k;
}
cut();
/*FOR(i,1,n+1) cout << a[i] << " ";
cout << "\n";*/
}
}
void ad(int x) {
nor();
int hei = 0;
FOR(i,x,x+k) hei = max(hei,a[i]);
hei ++;
FOR(i,1,n+1) if (i < x || i >= x+k) {
while (a[i] < hei) {
a[i] += k;
op.pb({1,i});
}
a[i] --;
}
op.pb({2,x});
cut();
}
int main() {
ios_base::sync_with_stdio(0); cin.tie(0);
cin >> n >> k;
FOR(i,1,n+1) {
cin >> a[i];
tmp[i] = a[i] % k;
}
FOR(i,2,n+1) while (tmp[i] != tmp[1]) {
if (i > n-k+1) {
cout << -1;
exit(0);
}
dif[i] ++;
FOR(j,i,i+k) tmp[j] = (tmp[j]+1)%k;
}
FOR(i,1,n+1) while (dif[i]) {
ad(i);
dif[i] --;
}
/*FOR(i,1,n+1) cout << a[i] << " ";
cout << "\n";*/
cout << sz(op) << "\n";
for (auto a: op) cout << a.f << " " << a.s << "\n";
}
/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
* if you have no idea just guess the appropriate well-known algo instead of doing nothing :/
*/
# |
결과 |
실행 시간 |
메모리 |
Grader output |
1 |
Correct |
3 ms |
248 KB |
Output is correct |
2 |
Correct |
2 ms |
360 KB |
Output is correct |
3 |
Correct |
2 ms |
400 KB |
Output is correct |
4 |
Incorrect |
2 ms |
512 KB |
Output isn't correct |
5 |
Halted |
0 ms |
0 KB |
- |
# |
결과 |
실행 시간 |
메모리 |
Grader output |
1 |
Incorrect |
3 ms |
632 KB |
Output isn't correct |
2 |
Halted |
0 ms |
0 KB |
- |
# |
결과 |
실행 시간 |
메모리 |
Grader output |
1 |
Correct |
3 ms |
248 KB |
Output is correct |
2 |
Correct |
2 ms |
360 KB |
Output is correct |
3 |
Correct |
2 ms |
400 KB |
Output is correct |
4 |
Incorrect |
2 ms |
512 KB |
Output isn't correct |
5 |
Halted |
0 ms |
0 KB |
- |
# |
결과 |
실행 시간 |
메모리 |
Grader output |
1 |
Correct |
3 ms |
248 KB |
Output is correct |
2 |
Correct |
2 ms |
360 KB |
Output is correct |
3 |
Correct |
2 ms |
400 KB |
Output is correct |
4 |
Incorrect |
2 ms |
512 KB |
Output isn't correct |
5 |
Halted |
0 ms |
0 KB |
- |