Da Da se divide en discusiones, y las ideas aún son muy inteligentes.
código:
#include <bits / stdc ++. h>
#define ll long long
#define lson now << 1
#define rson now << 1 | 1
#define N 100008
#define mod 998244353
#define setIO (s) freopen (s ".in "," r ", stdin)
usando el espacio de nombres std;
int n, m, cur;
ll f [N << 3], g [N << 3], SUM [N << 3], qp [N], mf [N << 3], mg [N << 3];
vacío markf (int ahora, int v)
{
f [ahora] = (ll) f [ahora] * qp [v]% mod;
SUMA [ahora] = (ll) SUMA [ahora] * qp [v]% mod;
mf [ahora] + = v;
}
vacío markg (int ahora, int v)
{
g [ahora] = (ll) g [ahora] * qp [v]% mod;
mg [ahora] + = v;
if (mf [ahora])
{
markf (lson, mf [ahora]);
markf (rson, mf [ahora]);
mf [ahora] = 0;
}
if (mg [ahora])
{
markg (lson, mg [ahora]);
markg (rson, mg [ahora]);
mg [ahora] = 0;
}
}
vacío pushup (int ahora)
{
SUM [ahora] = (ll) (SUM [lson] + SUM [rson] + f [ahora])% mod;
}
construcción vacía (int l, int r, int ahora)
{
f [ahora] = 0, g [ahora] = 1;
si (l == r) devuelve;
int mid = (l + r) >> 1;
build (l, mid, lson), build (mid + 1, r, rson);
}
actualización nula (int l, int r, int now, int L, int R)
{
pushdown (ahora);
if (l> = L && r <= R)
pushdown (p);
{
// 1 类 点
f [ahora] = (ll) (f [ahora] + qp [cur-1])% mod;
markf (lson, 1), markf (rson, 1);
pushup (ahora);
regreso;
}
int mid = (l + r) >> 1;
if (L <= mid && R> mid)
{
actualizar (l, mid, lson, L, R);
actualización (mid + 1, r, rson, L, R);
}
else if (L <= mid)
{
update (l, mid, lson, L, R);
int p = rson;
markf (p << 1,1), markf (p << 1 | 1,1);
markg (p << 1,1), markg (p << 1 | 1,1);
g [p] = (ll) (g [p] + g [p])% mod;
flexión (p);
}
else
{
actualización (mid + 1, r, rson, L, R);
int p = lson;
pushdown (p);
markf (p << 1,1), markf (p << 1 | 1,1);
markg (p << 1,1), markg (p << 1 | 1,1);
f [p] = (ll) (f [p] + qp [cur-1] -g [p] + mod)% mod;
g [p] = (ll) (g [p] + g [p])% mod;
flexión (p);
}
// 2 类 点
//
g [ahora] = (ll) (g [ahora] + qp [cur-1])% mod;
pushup (ahora);
}
int main ()
{
// setIO ("input");
qp [0] = 1;
para (int i = 1; i <N; ++ i) qp [i] = (ll) qp [i-1] * 2% mod;
scanf ("% d% d", & n, & m);
construir (1, n, 1);
para (int i = 1; i <= m; ++ i)
{
int op, l, r;
scanf ("% d", & op);
if (op == 1)
{
scanf ("% d% d", & l, & r), ++ cur, update (1, n, 1, l, r);
}
else
{
printf ("% lld \ n", SUM [1]);
}
}
devuelve 0;
}