Educational Codeforces Round (rated for Div. 2) D. degree Set

D. degree Set time limit per test 2 seconds memory limit per test megabytes input standard input output standard OUTPU T

You are given a sequence of n positive integers d1, d2, ..., dn (D1 < D2 < ... < DN). Your task is to construct a undirected graph such that:there is exactly DN + 1 vertices; There is no self-loops; There is no multiple edges; There is no more than 106 edges; Its degree set are equal to D.

Vertices should be numbered 1 through (DN + 1).

Degree sequence is an array a with length equal to the number of vertices in a graph such this AI is the number of Vertice s adjacent to i-th Vertex.

Degree set is a sorted in increasing order sequence of all distinct values from the degree sequence.

It is guaranteed this there exists such a graph that all the conditions hold, and it contains no more than 106 edges.

Print the resulting graph. Input

The first line contains one integer n (1≤n≤300)-the size of the degree set.

The second line contains n integers d1, d2, ..., dn (1≤di≤1000, D1 < D2 < ... < DN)-the degree set. Output

In the first line print one integer m (1≤m≤106)-the number of edges in the resulting graph. It is guaranteed this there exists such a graph that all the conditions hold and it contains no more than 106 edges.

Each of the next m lines should contain the integers vi and UI (1≤VI, UI≤DN + 1)-the description of the i-th edge. Examples input Copy

3
2 3 4

Output Copy

8
3 1
4 2
4 5
2 5
5 1
3 2
2 1
5 3

Input Copy

3
1 2 3

Output Copy

4
1 2
1 3
1 4
2 3

Test instructions: Constructs a diagram of a d[n]+1 node that requires degrees to be included in {D[1]...d[n]} and appears at least once.

Principle:

Prove:

#include <bits/stdc++.h> #include <ext/hash_map> #include <ext/hash_set> #include <ext/pb_ds/
assoc_container.hpp> #include <ext/pb_ds/tree_policy.hpp> #include <ext/pb_ds/priority_queue.hpp>
using namespace Std;
using namespace __gnu_cxx;
using namespace __gnu_pbds; #define Xinf Int_max #define INF 0x3f3f3f3f #define MP (x, y) make_pair #define PB (x) push_back (×) #define REP (X,n) for (int x=0; x<n; x + +) #define REP2 (X,L,R) for (int x=l; x<=r; x + +) #define DEP (x,r,l) for (int x=r; x>=l;
x--) #define CLR (a,x) memset (a,x,sizeof (A)) #define IT iterator #define RIT reverse_iterator typedef long Long LL;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef vector<pii> VII;
typedef vector<int> VI;
typedef TREE&LT;PII, Null_type, Greater<pii>, Rb_tree_tag, Tree_order_statistics_node_update > Rb_tree_set;
typedef TREE&LT;PII, PII, Greater<pii>, Rb_tree_tag, Tree_order_statistics_node_update > Rb_tree; #defiNE PQ std::p riority_queue #define HEAP __gnu_pbds::p riority_queue #define X First #define Y second #define LSON (x) ((x) &lt
; <1) #define Rson (x) ((x) <<1|1) #define EPS 10e-6 #define PI ACOs ( -1) #define N + int d[n], in[n];
VII G;
int n;
int S, t;
        void Solve (int n, int d[], int s) {REP (I, d[0]) {for (int j = s; j < T; ++j) G.PB (MP (J, T));
    t--;
        } if (n > 2) {REP2 (i, 1, n-1) {d[i] = d[i]-d[0];
    } Solve (N-2, d+1, t-d[n-2]);
    }} int main () {scanf ("%d", &n);
    REP (i, N) scanf ("%d", &d[i]);
    s = 1, t = d[n-1]+1;
    Solve (n, D, s);
    printf ("%d\n", G.size ());
    for (auto p:g) {printf ("%d%d\n", p.x, P.Y);
} return 0;
 }