【线段树 矩阵乘法dp】8.rseq

题目分析

#include<bits/stdc++.h>
#define MO 998244353
const int maxn = 200035;

struct Matrix
{
    int a[3][3];
    void init(int c, int spe)
    {
//        printf("spe:%d
",spe);
        a[0][0] = c, a[1][0] = 0, a[2][0] = 1ll*spe*c%MO;
        a[0][1] = 1, a[1][1] = 2, a[2][1] = 0;
        a[0][2] = 0, a[1][2] = 0, a[2][2] = 1;
    }
}f[maxn<<2];
int n,m;

int read()
{
    char ch = getchar();
    int num = 0, fl = 1;
    for (; !isdigit(ch); ch=getchar())
        if (ch=='-') fl = -1;
    for (; isdigit(ch); ch=getchar())
        num = (num<<1)+(num<<3)+ch-48;
    return num*fl;
}
int qmi(int a, int b)
{
    if (b <= -1) return 1;
    int ret = 1;
    for (; b; b>>=1,a=1ll*a*a%MO)
        if (b&1) ret = 1ll*ret*a%MO;
    return ret;
}
void debug(Matrix t)
{
    puts("------------------------------------");
    for (int i=0; i<3; i++, puts(""))
        for (int j=0; j<3; j++) printf("%d ",t.a[i][j]);
    puts("------------------------------------");
}
Matrix mult(Matrix a, Matrix b)
{
    Matrix ret;
    ret.a[0][0] = 0, ret.a[1][0] = 0, ret.a[2][0] = 0;
    ret.a[0][1] = 0, ret.a[1][1] = 0, ret.a[2][1] = 0;
    ret.a[0][2] = 0, ret.a[1][2] = 0, ret.a[2][2] = 0;
//    debug(a);
//    debug(b);
    for (int k=0; k<3; k++)
        for (int i=0; i<3; i++)
            for (int j=0; j<3; 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<<1], f[rt<<1|1]);
}
void build(int rt, int l, int r)
{
    if (l==r) f[rt].init(read(), qmi(2, l-2));
    else{
        int mid = (l+r)>>1;
        build(rt<<1, l, mid);
        build(rt<<1|1, mid+1, r);
        pushup(rt);
    }
}
void modify(int rt, int l, int r, int c, int w)
{
    if (l==r) f[rt].init(w, qmi(2, l-2));
    else{
        int mid = (l+r)>>1;
        if (c <= mid) modify(rt<<1, l, mid, c, w);
        else modify(rt<<1|1, mid+1, r, c, w);
        pushup(rt);
    }
}
void calc()
{
    Matrix ans;
    ans.a[0][0] = 0, ans.a[1][0] = 0, ans.a[2][0] = 0;
    ans.a[0][1] = 0, ans.a[1][1] = 0, ans.a[2][1] = 0;
    ans.a[0][2] = 1, ans.a[1][2] = 0, ans.a[2][2] = 0;
    ans = mult(ans, f[1]);
    printf("%d
",(ans.a[0][0]+ans.a[0][1])%MO);
}
int main()
{
    freopen("rseq.in","r",stdin);
    freopen("rseq.out","w",stdout);
    n = read(), m = read();
    build(1, 1, n), calc();
    for (int i=1; i<=m; i++)
    {
        int pos = read(), val = read();
        modify(1, 1, n, pos, val), calc();
    }
    return 0;
}