xdoj1429(二次剩余+FFT)

题目链接

http://acm.xidian.edu.cn/problem.php?id=1429

题解

好像有个技巧叫做二次剩余,以后再学,看公式还是能懂

发现$f(i)=\sum_{j=1}^{i}\lfloor\frac{n}{j} \rfloor\sqrt2^{j^2}(\sqrt2)^{-(i-j)^2}$ 是个卷积,直接上 $FFT$ 就可以了,然后 $\sqrt2$ 直接暴力枚举找模意义下的值就可以,复杂度为 $O(nlogn)$




代码

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/**
*         ┏┓    ┏┓
*         ┏┛┗━━━━━━━┛┗━━━┓
*         ┃       ┃  
*         ┃   ━    ┃
*         ┃ >   < ┃
*         ┃       ┃
*         ┃... ⌒ ...  ┃
*         ┃ ┃
*         ┗━┓ ┏━┛
*          ┃ ┃ Code is far away from bug with the animal protecting          
*          ┃ ┃ 神兽保佑,代码无bug
*          ┃ ┃           
*          ┃ ┃       
*          ┃ ┃
*          ┃ ┃           
*          ┃ ┗━━━┓
*          ┃ ┣┓
*          ┃ ┏┛
*          ┗┓┓┏━━━━━━━━┳┓┏┛
*           ┃┫┫ ┃┫┫
*           ┗┻┛ ┗┻┛
*/

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<stack>
#include<cmath>
#include<set>
#include<bitset>
#include<complex>
#include<assert.h>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define ll long long
#define eps 1e-8
#define succ(x) (1<<x)
#define mid (x+y>>1)
#define lowbit(x) (x&(-x))
#define sqr(x) ((x)*(x))
#define NM 1000005
#define nm 105
using namespace std;
const double pi=acos(-1);
const ll inf=998244353;
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;
}



int n;
ll t,a[NM],b[NM],ans;

ll qpow(ll x,ll t){return t?qpow(sqr(x)%inf,t>>1)*(t&1?x:1ll)%inf:1ll;}

struct FFT{
int n,bit,rev[NM];
ll b[NM],invn;
void fft(ll*a,int f){
inc(i,0,n-1)if(i<rev[i])swap(a[i],a[rev[i]]);
for(int k=1;k<n;k<<=1){
ll t=qpow(3,(inf-1)/k/2);if(f==-1)t=qpow(t,inf-2);
for(int i=0;i<n;i+=k<<1){
ll w=1;
for(int j=0;j<k;j++,w=w*t%inf){
ll x=a[i+j],y=w*a[i+j+k]%inf;
a[i+j]=(x+y)%inf;a[i+j+k]=(x-y+inf)%inf;
}
}
}
}
int plu(ll*a,ll*_b,int p,int m){
inc(i,0,m)b[i]=_b[i];
for(n=p+m+1,bit=0;succ(bit)<n;bit++);n=succ(bit);
invn=qpow(n,inf-2);
inc(i,0,n-1)rev[i]=(rev[i>>1]>>1)|((i&1)<<(bit-1));
fft(a,1);fft(b,1);inc(i,0,n-1)a[i]=a[i]*b[i]%inf;
fft(a,-1);inc(i,0,n-1)a[i]=a[i]*invn%inf;
inc(i,0,n-1)b[i]=0;
return p+m;
}
}fft;


int main(){
t=116195171;
int _=read();while(_--){
n=read();ans=0;
mem(a);mem(b);
inc(i,0,n)a[i]=qpow(qpow(t,1ll*i*i%(inf-1)),inf-2);
inc(i,1,n)b[i]=n/i*qpow(t,1ll*i*i%(inf-1))%inf;
fft.plu(a,b,n,n);
inc(i,1,n)ans+=b[i]*a[i]%inf,ans%=inf;
ans<<=1;ans%=inf;
inc(i,1,n)ans+=inf-qpow(2,1ll*i*i%(inf-1))*(n/i)%inf*(n/i)%inf,ans%=inf;
printf("%lld\n",ans);
}
return 0;
}