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,会存在一个,使得个还没有用过的最小的数的和为。那么当前的答案就是。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;
}