题解 fft
策略是每次取最高位
（赛时推出的策略是最高为 11 拆最高位，最高为 10 拆最高位的下一位，但仔细思考会发现这个其实等价于每次取最高位）
那么尝试写出贡献
最低位需要特判不好维护，考虑当成普通位来算再把算多的减掉
那么若第 \(i\) 位为 1，其贡献为
\[(i+1)2^{i+1}+\frac{i(i+1)}{2}2^i+2^i
\]
化简得
\[(i^2+5i+6)2^{i-1}
\]
然后 \(\sum i^22^{i-1}, \sum i2^{i-1}\) 都可以移位相减求出封闭形式
那么这个东西的前缀和可以 \(O(\log n)\) 求出
那么区间和也可以求了
那么珂朵莉树维护全 1 区间即可
复杂度 \(O(n\log n)\) 吧应该是
点击查看代码
// ubsan: undefined
// accoders
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 150010
#define fir first
#define sec second
#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 k, m;
const ll mod=998244353, inv2=(mod+1)>>1;
inline void md(ll& a, ll b) {a+=b; a=a>=mod?a-mod:a;}
inline ll qpow(ll a, ll b) {ll ans=1; for (; b; a=a*a%mod,b>>=1) if (b&1) ans=ans*a%mod; return ans;}
inline ll qsum(ll n) {n%=mod; return n*(n+1)%mod*inv2%mod;}
namespace force{
ll s;
int suf[N];
void solve() {
for (int i=1,l,r; i<=k; ++i) {
l=read(); r=read();
for (int j=l; j<=r; ++j) s+=1ll<<j;
}
for (int i=1,op,b; i<=m; ++i) {
op=read(); b=read();
if (op&1) s-=1ll<<b;
else s+=1ll<<b;
// cout<<"n: "<<s<<endl;
cout<<"s: "<<bitset<30>(s)<<endl;
ll t=s, ans=s;
bool fir=1;
suf[0]=t&1;
for (int i=1; i<64; ++i) suf[i]=suf[i-1]+(t&(1ll<<i)?1:0);
for (int i=63; i; --i) if (t&(1ll<<i)) {
if (fir) {
if (suf[i]==1) ans=(ans+(1ll<<i)%mod*i)%mod;
else ans=(ans+(1ll<<i+1)%mod*(i+1))%mod;
fir=0;
}
if (t&(1ll<<i-1)) {
ans=(ans+(1ll<<i)%mod*qsum(i))%mod;
if (suf[i-1]==1) ans=(ans+(1ll<<i-1)%mod*(i-1))%mod; //, cout<<"add2: "<<(1ll<<i-1)<<endl;
else ans=(ans+(1ll<<i)%mod*i)%mod; //, cout<<"add2: "<<(1ll<<i)<<endl;
// cout<<"add: "<<(1ll<<i)<<' '<<(1ll<<i)%mod*qsum(i)<<endl;
}
else {
t|=1ll<<i-1;
ans=(ans+(1ll<<i-1)%mod*qsum(i-1))%mod;
if (suf[i-1]==0) ans=(ans+(1ll<<i-1)%mod*(i-1))%mod; //, cout<<"add2: "<<(1ll<<i-1)<<endl;
else ans=(ans+(1ll<<i)%mod*i)%mod; //, cout<<"add2: "<<(1ll<<i)<<endl;
// cout<<"add: "<<(1ll<<i-1)<<' '<<(1ll<<i-1)%mod*qsum(i-1)<<endl;
}
}
printf("%lld\n", (ans%mod+mod)%mod);
}
}
}
namespace task1{
bool s[N];
set<int> st;
ll pw[N], dat[N], dat2[N], val;
void solve() {
pw[0]=1;
for (int i=1; i<=10005; ++i) pw[i]=(pw[i-1]<<1)%mod, dat[i]=pw[i]*i%mod, dat2[i]=pw[i]*qsum(i)%mod;
for (int i=1,l,r; i<=k; ++i) {
l=read(); r=read();
for (int j=l; j<=r; ++j) s[j]=1, val=(val+pw[j])%mod, st.insert(j);
}
for (int i=1,op,b; i<=m; ++i) {
op=read(); b=read();
if (op&1) {
val=((val-pw[b])%mod+mod)%mod;
int pos;
for (pos=b; !s[pos]; ++pos) s[pos]=1, st.insert(pos);
s[pos]=0, st.erase(pos);
}
else {
val=(val+pw[b])%mod;
int pos;
for (pos=b; s[pos]; ++pos) s[pos]=0, st.erase(pos);
s[pos]=1, st.insert(pos);
}
// cout<<"s: "; for (int i=*st.rbegin(); ~i; --i) {
// cout<<s[i];
// if (s[i]) assert(st.find(i)!=st.end());
// else assert(st.find(i)==st.end());
// } cout<<endl;
ll ans=val;
int high=*st.rbegin(), low=*st.begin();
if (high==low) md(ans, dat[high]);
else md(ans, dat[high+1]);
for (int i=high,lst=0; i; --i) if (s[i]||lst) {
lst=0;
if (s[i-1]) {
md(ans, dat2[i]);
if (i-1==low) md(ans, dat[i-1]);
else md(ans, dat[i]);
}
else {
lst=1;
md(ans, dat2[i-1]);
if (i-1<low) md(ans, dat[i-1]);
else md(ans, dat[i]);
}
}
printf("%lld\n", (ans%mod+mod)%mod);
}
}
}
namespace task{
ll ans, now;
struct node{mutable ll l, r, val;};
inline bool operator < (node a, node b) {return a.l<b.l;}
set<node> s;
inline ll sum1(ll r) {return (now-1)%mod;}
inline ll force_sum1(ll r) {ll ans=0; for (int i=1; i<=r; ++i) ans=(ans+qpow(2, i-1))%mod; return ans;}
inline ll sum2(ll r) {return (r*now-sum1(r))%mod;}
inline ll force_sum2(ll r) {ll ans=0; for (int i=1; i<=r; ++i) ans=(ans+i*qpow(2, i-1))%mod; return ans;}
inline ll sum3(ll r) {return (r*r%mod*now+sum1(r)-2*sum2(r))%mod;}
inline ll force_sum3(ll r) {ll ans=0; for (int i=1; i<=r; ++i) ans=(ans+i*i%mod*qpow(2, i-1))%mod; return ans;}
inline ll qsum(ll r) {if (r<=0) return 0; now=qpow(2, r); return (sum3(r)+5*sum2(r)+6*sum1(r))%mod;}
inline ll qval(int l, int r) {return (qsum(r)-qsum(l-1))%mod;}
int higher_1(int x) {
// cout<<"x: "<<x<<endl;
auto it=s.upper_bound({x, 0, 0});
if (it==s.begin()) return it->l;
if ((--it)->r<x) return (++it)->l;
return x;
}
int higher_0(int x) {
auto it=s.upper_bound({x, 0, 0});
if (it==s.begin()) return x;
if ((--it)->r<x) return x;
for (auto t=it; (++t)!=s.end()&&t->l==it->r+1; it=t);
return it->r+1;
}
auto split(int x) {
auto it=s.upper_bound({x, 0, 0});
if (it==s.begin()) return it;
if ((--it)->r<x) return ++it;
if (it->l==x) return it;
int l=it->l, r=it->r;
s.erase(it);
s.insert({l, x-1, qval(l, x-1)});
return s.insert({x, r, qval(x, r)}).fir;
}
void erase(int l, int r) {
if (l>r) return ;
auto it2=split(r+1), it1=split(l);
// s.erase(it1, it2);
for (; it1!=it2; ans=(ans-it1->val)%mod,it1=s.erase(it1));
}
void assign(int l, int r) {
if (l>r) return ;
auto it2=split(r+1), it1=split(l);
// s.erase(it1, it2);
for (; it1!=it2; ans=(ans-it1->val)%mod,it1=s.erase(it1));
s.insert({l, r, qval(l, r)});
ans=(ans+qval(l, r))%mod;
}
void solve() {
for (int i=1,l,r; i<=k; ++i) {
l=read(); r=read();
ans=(ans+qval(l, r))%mod;
s.insert({l, r, qval(l, r)});
}
for (int i=1,op,b; i<=m; ++i) {
// cout<<"i: "<<i<<endl;
op=read(); b=read();
if (op&1) {
int pos=higher_1(b);
// cout<<"pos: "<<pos<<endl;
erase(pos, pos), assign(b, pos-1);
}
else {
int pos=higher_0(b);
assign(pos, pos), erase(b, pos-1);
}
// cout<<"s: "; for (auto it:s) cout<<"("<<it.l<<','<<it.r<<") "; cout<<endl;
ll ctz=s.begin()->l;
if (ctz) printf("%lld\n", ((ans-(ctz+1)*qpow(2, ctz+1))%mod+mod)%mod);
else printf("%lld\n", ((ans+1)%mod+mod)%mod);
}
}
}
signed main()
{
freopen("fft.in", "r", stdin);
freopen("fft.out", "w", stdout);
k=read(); m=read();
// force::solve();
// task1::solve();
task::solve();
return 0;
}
