题目链接
题意
给定一个长度为(n)序列(a)，需要支持如下操作
单点修改
对于区间([l..r])，求(sum_{i=0}^{r-l}f_i*a_{l+i})，模(10^9)，其中(f)是斐波那契数列，(f_0=f_1=1)
区间加
做法
建线段树，对于一个节点([l..r])，记
[s_k=sum_{i=0}^{r-l}f_{i+k}*a_{l+i}
]
存在以下性质
[egin{align}
&s_{k-2}+s_{k-1}\
=&sum_{i=0}^{r-l}(f_{i+k-2}+f_{i+k-1})*a_{l+i}\
=&sum_{i=0}^{r-l}f_{i+k}*a_{l+i}\
=&s_k
end{align}
]
只需要维护(s_0)和(s_1)，预处理斐波那契数列作为系数，可以(O(1))求出(s_k)
合并两个区间的时候，需要求出第二个区间的(s_{len},s_{len+1})，其中(len)是第一个区间的长度
区间加，可以利用$$sum_{i=0}^nf_i=f_{n+2}-1$$快速修改
#include<cstdio>
#include<algorithm>
#include<ctype.h>
#include<string.h>
#include<math.h>
using namespace std;
#define ll long long
inline char read() {
static const int IN_LEN = 1000000;
static char buf[IN_LEN], *s, *t;
return (s == t ? t = (s = buf) + fread(buf, 1, IN_LEN, stdin), (s == t ? -1 : *s++) : *s++);
}
template<class T>
inline void read(T &x) {
static bool iosig;
static char c;
for (iosig = false, c = read(); !isdigit(c); c = read()) {
if (c == '-') iosig = true;
if (c == -1) return;
}
for (x = 0; isdigit(c); c = read()) x = ((x + (x << 2)) << 1) + (c ^ '0');
if (iosig) x = -x;
}
const int OUT_LEN = 10000000;
char obuf[OUT_LEN], *ooh = obuf;
inline void print(char c) {
if (ooh == obuf + OUT_LEN) fwrite(obuf, 1, OUT_LEN, stdout), ooh = obuf;
*ooh++ = c;
}
template<class T>
inline void print(T x) {
static int buf[30], cnt;
if (x == 0) print('0');
else {
if (x < 0) print('-'), x = -x;
for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 + 48;
while (cnt) print((char)buf[cnt--]);
}
}
inline void flush() { fwrite(obuf, 1, ooh - obuf, stdout); }
const int N = 200005, P = 1000000000;
int n, m, f[N], a[N], lazy[N<<2];
struct st{
int a, b, len;
inline st(){}
inline st(int a_, int b_, int len_){ a=a_, b=b_, len=len_;}
inline st operator *(const st &rhs)const{
return st((a+(ll)rhs.a*f[len-1]+(ll)rhs.b*f[len])%P, (b+(ll)rhs.a*f[len]+(ll)rhs.b*f[len+1])%P, len+rhs.len);
}
}s[N<<2];
void build(int l, int r, int t){
if(l==r) return (void)(s[t].a=s[t].b=a[l], s[t].len=1);
int mid=l+r>>1, k=t<<1;
build(l, mid, k), build(mid+1, r, k|1);
s[t]=s[k]*s[k|1];
}
inline void add(int t, int x){
(lazy[t]+=x)%=P;
s[t].a=(s[t].a+(ll)(f[s[t].len+2]-1)*x)%P;
s[t].b=(s[t].b+(ll)(f[s[t].len+3]-2)*x)%P;
}
void modify(int l, int r, int t, int x, int y){
if(l==r) return (void)(s[t].a=s[t].b=y);
int mid=l+r>>1, k=t<<1;
if(lazy[t]) add(k, lazy[t]), add(k|1, lazy[t]), lazy[t]=0;
if(x<=mid) modify(l, mid, k, x, y); else modify(mid+1, r, k|1, x, y);
s[t]=s[k]*s[k|1];
}
st query(int l, int r, int t, int L, int R){
if(L<=l && r<=R) return s[t];
int mid=l+r>>1, k=t<<1;
if(lazy[t]) add(k, lazy[t]), add(k|1, lazy[t]), lazy[t]=0;
if(R<=mid) return query(l, mid, k, L, R);
if(L>mid) return query(mid+1, r, k|1, L, R);
return query(l, mid, k, L, R)*query(mid+1, r, k|1, L, R);
}
void change(int l, int r, int t, int L, int R, int x){
if(L<=l && r<=R) return add(t, x);
int mid=l+r>>1, k=t<<1;
if(lazy[t]) add(k, lazy[t]), add(k|1, lazy[t]), lazy[t]=0;
if(L<=mid) change(l, mid, k, L, R, x);
if(R>mid) change(mid+1, r, k|1, L, R, x);
s[t]=s[k]*s[k|1];
}
int main() {
read(n), read(m);
f[1]=f[2]=1;
for(int i=3; i<=n+3; ++i) f[i]=(f[i-1]+f[i-2])%P;
for(int i=1; i<=n; ++i) read(a[i]);
build(1, n, 1);
while(m--){
static int opt, l, r, x;
read(opt), read(l), read(r);
if(opt==1) modify(1, n, 1, l, r);
else if(opt==2) print(query(1, n, 1, l, r).a), print('
');
else read(x), change(1, n, 1, l, r, x);
}
return flush(), 0;
}