$$\\sum_{i=0}^{2^{n}}{lowbit(i)} $$

求
\[\sum_{i=0}^{2^{n}}{lowbit(i)} \]
$(n≤10^{18})$答案对998244353取模

lowbit

首先数组数组里经常出现的lowit函数
对于lowbit

给定非负整数n。记lowbit(x)为x的二进制表示下最低位的1所对应的值
比如
lowbit(1),1的二进制位1,最低为对应的是1
lowbit(3),3的二进制为11,最低位对应的是1
lowbit(4),4的二进制为100,最低位对应的是4

一般来说2^n的lowbit为本身，奇数的lowbit为1

比如n=2
$\sum_{i=0}^{2^{n}}{lowbit(i)}\\ = \sum_{i=0}^{4}{lowbit(i)}\\ = lowbit(0) + lowbit(1)+ lowbit(2)+ lowbit(3)+ lowbit(4)$

而对于$lowbit(0)=0,lowbit(2^{n})=2^{n}$
而对于$2^{n}有2^{n-1}个奇数$
所以式子变成了求
$lowbit(2) + lowbit(4) +lowbit(6).....+lowbit(2^{n-1})+2^{n}+2^{n-1}$
$令x=lowbit(2) + lowbit(4) +lowbit(6).....+lowbit(2^{n-1})$

打表

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <vector>
#define ll long long
using namespace std;
const int mod = 998244353;
const int maxn = 1e5+5;
const int inf = 0x3f3f3f3f;
ll read(){
    ll x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch)){x=x*10+ch-48;ch=getchar();}
    return f*x;
}
ll lowbit(ll x){
    return (x&(-x)) % mod;
}
ll quickpow(ll a,ll b){
    ll ans = 1;
    while(b){
        if(b&1) ans = ans * a % mod;
        b >>= 1;
        a = a * a%mod; 
    }
    return ans;
}
void debug(){
    for(ll n = 1; n <= 20; n++){
        ll all = quickpow(2,n);
        ll ans = 0;
        for(ll j = 1;j <= all; j++){
            ans = (ans + lowbit(j)) % mod;
        }
        cout<<n<<" "<<ans<<" "<<ans-all-all/2<<endl;
    }
}
int main(){
    debug();
    return 0;
}

打表发现规律

n  x
1  0
2  2
3  8
4  24
5  64
6  160
7  384
8  896
9  2048
10  4608
11  10240
12  22528
13  49152
14  106496
15  229376
16  491520
17  1048576
18  2228224
19  4718592
20  9961472

推导

发现$a_{n+1}=2a_{n}+2^{n}$
所以理由生成函数解
$a_{n+1}-2a_{n} = 2^{n}$
①求齐次
$a_{n+1}-2a_{n} = 0$
$ x-2=0 $解得$x=2$，所以通解为$c_{1}2^{n}$
②求非齐次
$x=2$为1重根且等于常数，所以特解为$Pn2^{n}$

代入非齐次$P(n+1)2^{n+1}=2Pn2^{n}+2^{n}$
解得$P=\frac{1}{2}$
特解为$n2^{n-1}$
③求解
解为齐次通解+非齐次特解
$x =c_{1}2^{n}+ n2^{n-1} ，其中n=1时，x=0$
$c_{1}=-\frac{1}{2}$
$x = (n-1)2^{n-1}$

与上面合并得到
$a_{n} = (n-1)2^{n-1}+2^{n}+2^{n-1} = (n+2)2^{n-1}$
$∴a_{n} = (n+2)2^{n-1}$

代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#define ll long long
#define mod 998244353
using namespace std;
const int maxn = 1e5+5;
const int inf = 0x3f3f3f3f;
ll read(){
    ll x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch)){x=x*10+ch-'0';ch=getchar();}
    return f*x;
}
ll pow(ll a,ll b,ll p){
    ll ans = 1;
    while(b){
        if(b&1)ans = ans * a % p;
        b >>= 1;
        a = a * a %p;
    }
    return ans;
}
void getans(ll n){
    if(n == 0)cout<<"1"<<endl;
    else{
        ll ans = ((n + 2) % mod) * (pow(2,n-1,mod) %mod) %mod;
        cout<<ans<<endl;
    }
}
int main(){
    ll t = read();
    while(t--){
        ll n = read();
        getans(n);
    }
    return 0;
}

来源：https://www.cnblogs.com/Emcikem/p/11877270.html

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