Thallium54

Codeforces 1208D- Restore Permutation 题解

二分 数据结构 树状数组 题解

题目本身就很好，同时又能带来对树状数组的一些思考。

题解 #

我们要倒着处理，对于当前的ii，会存在一个kk，使得kk个还没有用过的最小的数的和为sis_i。那么当前ii的答案就是k+1k+1。可以用树状数组配二分找，也可以用树状数组配倍增黑科技求。

Code #

二分 #

#include <bits/stdc++.h>
 
#define forn(i, n) for (int i = 0; i < int(n); ++i)
#define for1(i, n) for (int i = 1; i <= int(n); ++i)
#define fore(i, l, r) for (int i = int(l); i <= int(r); ++i)
#define ford(i, n) for (int i = int(n)-1; i >= 0; --i)
#define pb push_back
#define eb emplace_back
#define ms(a, x) memset(a, x, sizeof(a))
#define F first
#define S second
#define endl '\n'
#define all(x) (x).begin(),(x).end()
 
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int INF = 0x3f3f3f3f;
mt19937 gen(chrono::high_resolution_clock::now().time_since_epoch().count());
template<typename... Args>
void write(Args... args) { ((cout << args << " "), ...); cout<<endl;}
 
struct fenwick{
    vector<ll> t;
    int n;
    fenwick(int n_):n(n_){
        t=vector<ll>(n+1);
    }
    void update(int i,int x){
        for(;i<=n;i+=i&-i){
            t[i]+=x;
        }
    }
    ll get(int i){
        ll res=0;
        for(;i>0;i-=i&-i) res+=t[i];
        return res;
    }
};
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    cin>>n;
    vector<ll> a(n);
    for(auto& it:a) cin>>it;
    fenwick tree(n);
    for(int i=1;i<=n;i++) tree.update(i,i);
    vector<int> ans(n);
    for(int i=n-1;i>=0;i--){
        int l=1,r=n;
        while(l<=r){
            int mid=(l+r)>>1;
            if(tree.get(mid)<=a[i]) l=mid+1;
            else r=mid-1;
        }
        ans[i]=l;
        tree.update(l,-l);
    }
    for(auto it:ans) cout<<it<<' ';
    return 0;
}

倍增 #

#include <bits/stdc++.h>
 
#define forn(i, n) for (int i = 0; i < int(n); ++i)
#define for1(i, n) for (int i = 1; i <= int(n); ++i)
#define fore(i, l, r) for (int i = int(l); i <= int(r); ++i)
#define ford(i, n) for (int i = int(n)-1; i >= 0; --i)
#define pb push_back
#define eb emplace_back
#define ms(a, x) memset(a, x, sizeof(a))
#define F first
#define S second
#define endl '\n'
#define all(x) (x).begin(),(x).end()
 
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int INF = 0x3f3f3f3f;
mt19937 gen(chrono::high_resolution_clock::now().time_since_epoch().count());
template<typename... Args>
void write(Args... args) { ((cout << args << " "), ...); cout<<endl;}
 
struct fenwick{
    vector<ll> t;
    int n;
    fenwick(int n_):n(n_){
        t=vector<ll>(n+1);
    }
    void update(int i,int x){
        for(;i<=n;i+=i&-i){
            t[i]+=x;
        }
    }
    int search(ll prefix){
        int pos=0;
        ll sum=0;
        for(int i=20;i>=0;i--){
            if(pos+(1<<i)<=n&&(sum+t[pos+(1<<i)]<=prefix)){
                pos+=(1<<i);
                sum+=t[pos];
            }
        }
        return pos+1;
    }
};
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    cin>>n;
    vector<ll> a(n);
    for(auto& it:a) cin>>it;
    fenwick tree(n);
    for(int i=1;i<=n;i++) tree.update(i,i);
    vector<int> ans(n);
    for(int i=n-1;i>=0;i--){
        int x=tree.search(a[i]);
        ans[i]=x;
        tree.update(x,-x);
    }
    for(auto it:ans) cout<<it<<' ';
    return 0;
}