【线段树 矩阵乘法dp】8.rseq

题目分析

#include<bits/stdc++.h>
#define MO 998244353
const int maxn = ;
 
struct Matrix
{
    int a[][];
    void init(int c, int spe)
    {
//        printf("spe:%d\n",spe);
        a[][] = c, a[][] = , a[][] = 1ll*spe*c%MO;
        a[][] = , a[][] = , a[][] = ;
        a[][] = , a[][] = , a[][] = ;
    }
}f[maxn<<];
int n,m;
 
int read()
{
    char ch = getchar();
    int num = , fl = ;
    for (; !isdigit(ch); ch=getchar())
        if (ch=='-') fl = -;
    for (; isdigit(ch); ch=getchar())
        num = (num<<)+(num<<)+ch-;
    return num*fl;
}
int qmi(int a, int b)
{
    if (b <= -) return ;
    int ret = ;
    for (; b; b>>=,a=1ll*a*a%MO)
        if (b&) ret = 1ll*ret*a%MO;
    return ret;
}
void debug(Matrix t)
{
    puts("------------------------------------");
    for (int i=; i<; i++, puts(""))
        for (int j=; j<; j++) printf("%d ",t.a[i][j]);
    puts("------------------------------------");
}
Matrix mult(Matrix a, Matrix b)
{
    Matrix ret;
    ret.a[][] = , ret.a[][] = , ret.a[][] = ;
    ret.a[][] = , ret.a[][] = , ret.a[][] = ;
    ret.a[][] = , ret.a[][] = , ret.a[][] = ;
//    debug(a);
//    debug(b);
    for (int k=; k<; k++)
        for (int i=; i<; i++)
            for (int j=; j<; j++)
                ret.a[i][j] = (ret.a[i][j]+1ll*a.a[i][k]*b.a[k][j]%MO)%MO;
//    debug(ret);
//    puts("@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@");
    return ret;
}
void pushup(int rt)
{
    f[rt] = mult(f[rt<<], f[rt<<|]);
}
void build(int rt, int l, int r)
{
    if (l==r) f[rt].init(read(), qmi(, l-));
    else{
        int mid = (l+r)>>;
        build(rt<<, l, mid);
        build(rt<<|, mid+, r);
        pushup(rt);
    }
}
void modify(int rt, int l, int r, int c, int w)
{
    if (l==r) f[rt].init(w, qmi(, l-));
    else{
        int mid = (l+r)>>;
        if (c <= mid) modify(rt<<, l, mid, c, w);
        else modify(rt<<|, mid+, r, c, w);
        pushup(rt);
    }
}
void calc()
{
    Matrix ans;
    ans.a[][] = , ans.a[][] = , ans.a[][] = ;
    ans.a[][] = , ans.a[][] = , ans.a[][] = ;
    ans.a[][] = , ans.a[][] = , ans.a[][] = ;
    ans = mult(ans, f[]);
    printf("%d\n",(ans.a[][]+ans.a[][])%MO);
}
int main()
{
    freopen("rseq.in","r",stdin);
    freopen("rseq.out","w",stdout);
    n = read(), m = read();
    build(, , n), calc();
    for (int i=; i<=m; i++)
    {
        int pos = read(), val = read();
        modify(, , n, pos, val), calc();
    }
    return ;
}