「ABC221E」LEQ 題解
E - LEQ
Time Limit: \(2\; sec\) / Memory Limit: \(1024\; MB\)
Score : \(500\; points\)
Problem Statement|題目描述
-
Given is a sequence of \(N\) integers: \(A=(A_1 ,A_2 ,…,A_N )\).
-
給定一個包含 \(N\) 個整數的序列: \(A=(A_1 ,A_2 ,…,A_N )\)。
-
Find the number of (not necessarily contiguous) subsequences$ A′ =(A_1′ ,A_2′ ,…,A_k′ )$ of length at least \(2\)
that satisfy the following condition:
\(A_1′\leq A_k′ .\)
- 求子序列的個數(不一定是連續的)\(A′ =(A_1′ ,A_2′ ,…,A_k′ )\)，要求長度至少為 \(2\)，且滿足以下條件：
\(A_1′\leq A_k′ .\)
-
Since the count can be enormous, print it modulo \(998244353\).
-
由於結果可能很大，所以將其對 \(998244353\) 取模輸出。
-
Here, two subsequences are distinguished when they originate from different sets of indices, even if they are the same as sequences.
-
在這裡，當兩個子序列來自不同的集合位置時，即使它們與序列相同，也可以區分它們。
Constraints|資料範圍
-
\(2\leq N\leq 3\times 10^5\)
-
\(1\leq A_i\leq 10^9\)
-
All values in input are integers.
-
\(2\leq N\leq 3\times 10^5\)
-
\(1\leq A_i\leq 10^9\)
-
輸入中的所有值都是整數。
Input|輸入
Input is given from Standard Input in the following format:
\(N\)
\(A_1\ A_2\ …\ A_N\)
- 輸入為以下格式的標準輸入（中間有空格）：
\(N\)
\(A_1\ A_2\ …\ A_N\)
Output|輸出
-
Print the number of (not necessarily contiguous) subsequences$ A′ =(A_1′ ,A_2′ ,…,A_k′ )$ of length at least \(2\) that satisfy the condition in Problem Statement.
-
輸出子序列的個數(不一定是連續的)\(A′ =(A_1′ ,A_2′ ,…,A_k′ )\)，要求長度至少為\(2\)，且滿足題目描述中的條件。
分析
AC程式碼
#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
const int mod=998244353;
const int maxn=300010;
const int inf=0x3f3f3f3f;
int n;
inline ll power(ll a,ll b){
ll ans=1;
while(b){
if(b&1)ans=(ans*a)%mod;
b>>=1;
a=(a*a)%mod;
}
return ans%mod;
}
inline ll inv(ll x){return power(x,mod-2);}//逆元
ll a[maxn],b[maxn];
int sum[maxn];
inline int lowbit(int x){return x&(-x);}
inline void modify(int x,int k){
while(x<=n){
sum[x]=(sum[x]+k)%mod;
x+=lowbit(x);
}
}
inline ll ask(int x){
ll res=0;
while(x){
res=(res+sum[x])%mod;
x-=lowbit(x);
}
return res;
}
inline ll query(int l,int r){
ll res=(ask(r)-ask(l-1)+mod)%mod;
// printf("%d\n",res);
return res;
}
int main(){
scanf("%d",&n);
for(int i=1;i<=n;i++){
scanf("%lld",&a[i]);
b[i]=a[i];
}
sort(b+1,b+1+n);
cc_hash_table<int,int>rank;//雜湊表
for(int i=1;i<=n;i++)rank[b[i]]=i;//離散化
for(int i=1;i<=n;i++)//樹狀陣列
modify(rank[a[i]],power(2,i));
ll ans=0;
for(int i=1;i<=n;i++){
modify(rank[a[i]],mod-power(2,i));
ans+=inv(power(2,i+1))*query(rank[a[i]],n);
ans%=mod;
}
printf("%lld",ans);
return 0;
}