[atARC103F]Distance Sums

给定$n$个数$d_{i}$，构造一棵$n$个点的树使得$forall 1le ile n,sum_{j=1}^{n}dist(i,j)=d_{i}$

其中$dist(i,j)$表示$i$到$j$的路径上所经过的边数，若无解输出-1

$2le nle 10^{5}$，$1le d_{i}le 10^{12}$，保证$d_{i}$各不相同

#include<bits/stdc++.h>
using namespace std;
#define N 100005
#define pli pair<long long,int>
#define fi first
#define se second
vector<pli>e;
int n,sz[N];
pli d[N];
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%lld",&d[i].fi);
        d[i].se=i;
    }
    sort(d+1,d+n+1);
    for(int i=1;i<=n;i++)sz[i]=1;
    for(int i=n;i>1;i--){
        int fa=lower_bound(d+1,d+i+1,make_pair(d[i].fi+(2*sz[i]-n),0))-d;
        if (d[fa].fi!=d[i].fi+(2*sz[i]-n)){
            printf("-1");
            return 0;
        }
        sz[fa]+=sz[i];
        e.push_back(make_pair(d[fa].se,d[i].se));
    }
    for(int i=2;i<=n;i++)d[1].fi-=sz[i];
    if (d[1].fi)printf("-1");
    else
        for(int i=0;i<n-1;i++)printf("%d %d
",e[i].fi,e[i].se);
}