제출 #70388

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
70388		Benq	JOIRIS (JOI16_joiris)	C++14	0 / 100	3 ms	816 KiB

joiris

#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], nex[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();
}

void genDif() {
    FOR(i,1,n-k+2) dif[i] = -1;
    FOR(i,2,n+1) {
        // dif[i]+a[i]-a[i-1] = dif[i-k]
        if (i-k < 1) {
            if (i > n-k+1) {
                if ((a[i]-a[i-1]) % k != 0) {
                    // cout << "A\n";
                    cout << -1;
                    exit(0);
                }
                continue;
            } else {
                dif[i] = ((a[i-1]-a[i])%k+k)%k;
            }
        } else {
            if (i > n-k+1) {
                if (dif[i-k] != -1 && dif[i-k] != ((a[i]-a[i-1])%k+k)%k) {
                    // cout << "B\n";
                    cout << -1;
                    exit(0);
                }
                dif[i-k] = ((a[i]-a[i-1])%k+k)%k;
            } else {
                nex[i-k] = ((a[i]-a[i-1])%k+k)%k; 
            }
        }
    }
    FOR(i,1,min(n-k+2,k+1)) {
        int lo = i, hi = i; while (hi+k <= n-k+1) hi += k;
        if (dif[lo] != -1) {
            for (int j = lo+k; j <= hi; j += k) {
                if (dif[j] > -1 && dif[j] != (dif[j-k]-nex[j-k]+k)%k) {
                    cout << -1;
                    exit(0);
                }
                dif[j] = (dif[j-k]-nex[j-k]+k)%k;
            }
        } else {
            if (dif[hi] == -1) dif[hi] = 0;
            for (int j = hi-k; j >= 0; j -= k) {
                if (dif[j] > -1 && dif[j] != (dif[j+k]+nex[j])%k) {
                    cout << -1;
                    exit(0);
                }
                dif[j] = (dif[j+k]+nex[j])%k;
            }
        }
    }
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> n >> k;
    FOR(i,1,n+1) cin >> a[i];
    genDif();
    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 :/
*/

• Subtask 1: 0 / 15
• Subtask 2: 0 / 15
• Subtask 3: 0 / 15
• Subtask 4: 0 / 55