题解 人生吧
看着比较可做，于是我反手一个莫比乌斯反演 T 成了 27pts
其实不需要反演
每个质因子及其幂次是独立的
\[ans=\prod\limits_p\prod\limits_i(p^i)^{\sum\limits_s[\gcd=p^i]}
\]
\[ans=\prod\limits_p\prod\limits_ip^{\sum\limits_s[\gcd\geqslant p^i]}
\]
\[ans=\prod\limits_p\prod\limits_ip^{2^{cnt_{p^i}}}
\]
\(cnt_{p^i}\) 为 \([l, r]\) 中 \(p_i\) 作为因子出现的次数
格式以后再修吧
那么直接莫队+预处理质数幂次、逆元可以做到 \(O(n\sqrt n\log n)\)
发现 \(\leqslant \sqrt n\) 的质数和幂次的组合数量很少，可以前缀和处理
然后 \(>\sqrt n\) 的质数再莫队就可以做到 \(O(\sqrt n)\)
来不及写了，放个带 log 的跑路
点击查看代码
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 100010
#define pb push_back
#define ll long long
//#define int long long
char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
int ans=0, f=1; char c=getchar();
while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
return ans*f;
}
int n, m;
int a[N];
const ll mod=998244353, mod2=mod-1;
inline int gcd(int a, int b) {return !b?a:gcd(b, a%b);}
inline ll qpow(ll a, ll b, ll mod=::mod) {ll ans=1; for (; b; a=a*a%mod,b>>=1) if (b&1) ans=ans*a%mod; return ans;}
namespace force{
int sta[N], top;
ll query(int l, int r) {
ll ans=1; top=0;
for (int i=l; i<=r; ++i) sta[top++]=a[i];
int lim=1<<top;
for (int s=1; s<lim; ++s) {
bool flag=0;
int tem=1;
for (int i=0; i<top; ++i) if (s&(1<<i)) {
if (!flag) tem=sta[i], flag=1;
else tem=gcd(tem, sta[i]);
}
// cout<<"tem: "<<tem<<endl;
ans=ans*tem%mod;
}
return ans;
}
void solve() {
for (int i=1,l,r; i<=m; ++i) {
l=read(); r=read();
printf("%lld\n", query(l, r));
}
}
}
namespace task1{
bool npri[N];
vector<int> div[N];
ll cnt[N], ans[N], now=1;
int pri[N], mu[N], bel[N], pcnt, sqr;
struct que{int l, r, id;}q[N];
inline bool operator < (que a, que b) {return bel[a.l]==bel[b.l]?((bel[a.l]&1)?a.r<b.r:a.r>b.r):bel[a.l]<bel[b.l];}
void add(int b) {
for (auto& g:div[b]) {
ll sum=0;
for (auto& d:div[b/g])
sum+=mu[d]*qpow(2, cnt[g*d], mod-1);
now=now*qpow(g, (sum%mod2+mod2)%mod2)%mod;
}
for (auto& it:div[b]) ++cnt[it];
}
void del(int b) {
for (auto& it:div[b]) --cnt[it];
for (auto& g:div[b]) {
ll sum=0;
for (auto& d:div[b/g])
sum+=mu[d]*qpow(2, cnt[g*d], mod-1);
now=now*qpow(qpow(g, (sum%mod2+mod2)%mod2), mod-2)%mod;
}
}
void solve() {
sqr=sqrt(n);
for (int i=1; i<=n; ++i) bel[i]=(i-1)/sqr+1;
for (int i=1; i<=1e5; ++i)
for (int j=i; j<=1e5; j+=i)
div[j].pb(i);
mu[1]=1;
for (int i=2; i<=1e5; ++i) {
if (!npri[i]) pri[++pcnt]=i, mu[i]=-1;
for (int j=1,x; j<=pcnt&&i*pri[j]<=1e5; ++j) {
npri[x=i*pri[j]]=1;
if (!(i%pri[j])) break;
else mu[x]=-mu[i];
}
}
for (int i=1,l,r; i<=m; ++i) {
l=read(); r=read();
q[i]={l, r, i};
}
sort(q+1, q+m+1);
for (int i=1,l=1,r=0; i<=m; ++i) {
// cout<<"i: "<<i<<endl;
while (l>q[i].l) add(a[--l]);
while (r<q[i].r) add(a[++r]);
while (l<q[i].l) del(a[l++]);
while (r>q[i].r) del(a[r--]);
ans[q[i].id]=now;
}
for (int i=1; i<=m; ++i) printf("%lld\n", ans[i]);
}
}
namespace task2{
bool npri[N];
vector<int> div[N];
int pri[N], mu[N], bel[N], pcnt, sqr;
ll cnt[N], ans[N], rec[N], pw[N], now=1;
struct que{int l, r, id;}q[N];
inline bool operator < (que a, que b) {return bel[a.l]==bel[b.l]?((bel[a.l]&1)?a.r<b.r:a.r>b.r):bel[a.l]<bel[b.l];}
void add(int b) {
for (auto& g:div[b]) {
ll sum=0;
for (auto& d:div[b/g])
sum+=mu[d]*pw[cnt[g*d]];
now=now*qpow(g, (sum%mod2+mod2)%mod2)%mod;
}
for (auto& it:div[b]) ++cnt[it];
}
void del(int b) {
for (auto& it:div[b]) --cnt[it];
for (auto& g:div[b]) {
ll sum=0;
for (auto& d:div[b/g])
sum+=mu[d]*pw[cnt[g*d]];
now=now*qpow(qpow(g, (sum%mod2+mod2)%mod2), mod-2)%mod;
}
}
void solve() {
sqr=sqrt(n);
for (int i=1; i<=n; ++i) bel[i]=(i-1)/sqr+1;
for (int i=1; i<=1e5; ++i)
for (int j=i; j<=1e5; j+=i)
div[j].pb(i);
// for (int i=1; i<=1e5; ++i) {
// cout<<i<<": "; for (auto it:div[i]) cout<<it<<' '; cout<<endl;
// } exit(0);
mu[1]=1;
for (int i=2; i<=1e5; ++i) {
if (!npri[i]) pri[++pcnt]=i, mu[i]=-1;
for (int j=1,x; j<=pcnt&&i*pri[j]<=1e5; ++j) {
npri[x=i*pri[j]]=1;
if (!(i%pri[j])) break;
else mu[x]=-mu[i];
}
}
pw[0]=1;
for (int i=1; i<=1e5; ++i) pw[i]=pw[i-1]*2%(mod-1);
for (int i=1,l,r; i<=m; ++i) {
l=read(); r=read();
q[i]={l, r, i};
}
sort(q+1, q+m+1);
for (int i=1,l=1,r=0; i<=m; ++i) {
// cout<<"i: "<<i<<' '<<l<<' '<<r<<endl;
while (l>q[i].l) add(a[--l]);
while (r<q[i].r) add(a[++r]);
while (l<q[i].l) del(a[l++]);
while (r>q[i].r) del(a[r--]);
ans[q[i].id]=now;
}
for (int i=1; i<=m; ++i) printf("%lld\n", ans[i]);
}
}
namespace task{
bool npri[N];
ll cnt[N], ans[N], now=1;
vector<ll> pw[N], inv[N];
int siz[N], pri[N], bel[N], low[N], lowp[N], pcnt, sqr;
struct que{int l, r, id;}q[N];
inline bool operator < (que a, que b) {return bel[a.l]==bel[b.l]?((bel[a.l]&1)?a.r<b.r:a.r>b.r):bel[a.l]<bel[b.l];}
void add(int val) {
for (int t=val; t>1; t/=low[t])
for (int p=low[t]; p>1; p/=lowp[p])
now=now*pw[lowp[p]][cnt[p]++]%mod;
}
void del(int val) {
for (int t=val; t>1; t/=low[t])
for (int p=low[t]; p>1; p/=lowp[p])
now=now*inv[lowp[p]][--cnt[p]]%mod;
}
void solve() {
sqr=sqrt(n);
for (int i=1; i<=n; ++i) bel[i]=(i-1)/sqr+1;
for (int i=2; i<=1e5; ++i) {
if (!npri[i]) pri[++pcnt]=low[i]=lowp[i]=i;
for (int j=1,x; j<=pcnt&&i*pri[j]<=1e5; ++j) {
npri[x=i*pri[j]]=1;
if (!(i%pri[j])) {low[x]=low[i]*pri[j]; lowp[x]=pri[j]; break;}
else low[x]=lowp[x]=pri[j];
}
}
for (int i=1; i<=n; ++i)
for (int t=a[i]; t>1; t/=low[t])
for (int p=low[t]; p>1; p/=lowp[p])
++siz[p];
for (int i=1; i<=1e5; ++i)
for (int t=i; t>1; t/=low[t])
for (int p=low[t]; p>1; p/=lowp[p])
siz[lowp[p]]=max(siz[lowp[p]], siz[p]);
for (int i=1; i<=1e5; ++i) if (!npri[i]) {
pw[i].resize(siz[i]+5); inv[i].resize(siz[i]+5);
for (int j=0; j<=siz[i]; ++j) pw[i][j]=qpow(i, qpow(2, j, mod-1));
for (int j=0; j<=siz[i]; ++j) inv[i][j]=qpow(qpow(i, qpow(2, j, mod-1)), mod-2);
}
for (int i=1,l,r; i<=m; ++i) {
l=read(); r=read();
q[i]={l, r, i};
}
sort(q+1, q+m+1);
for (int i=1,l=1,r=0; i<=m; ++i) {
// cout<<"i: "<<i<<' '<<l<<' '<<r<<endl;
while (l>q[i].l) add(a[--l]);
while (r<q[i].r) add(a[++r]);
while (l<q[i].l) del(a[l++]);
while (r>q[i].r) del(a[r--]);
ans[q[i].id]=now;
}
for (int i=1; i<=m; ++i) printf("%lld\n", ans[i]);
}
}
signed main()
{
freopen("C.in", "r", stdin);
freopen("C.out", "w", stdout);
n=read(); m=read();
for (int i=1; i<=n; ++i) a[i]=read();
// force::solve();
task::solve();
return 0;
}
