题目链接
不得不说这题是线段树维护矩阵的一道好题,此外推荐\(LibreOJ\)上的一道好题「THUSCH 2017」大魔法师 也可以用线段树维护矩阵
Solution [CF718C] Sasha and Array
题目大意:请你维护一个数列,支持一下两种操作:
\(1\).将区间\([l,r]\)内的数加上\(x\)
\(2\).求\(\sum_{i =l}^{r}f(a_{i})\),其中\(f(x)\)表示斐波那契数列的第\(x\)项
做法:既然数据范围已经达到了\(n,m \leq 1e5\)级别,那么我们就得考虑\(O(nlogn)\)级别的算法了
回想一下我们是如何以\(O(logn)\)的优秀复杂度求斐波那契数列的.矩阵乘法对吧?
我们设一个矩阵\(orgin = \begin{bmatrix} 0&1\end{bmatrix}\),以及一个转移矩阵\(w = \begin{bmatrix} 1&1\\1&0 \end{bmatrix}\)
那么很显然\(f(x) = orgin\;*\;w^x\) 即对于表示\(f(x)\)的矩阵乘上\(w^k\)就求得了\(f(x + k)\)
很显然我们已经解决了操作\(1\)了
对于操作\(2\),我们只需要维护区间矩阵之和:正确性显而易见,矩阵满足结合律.所以有:
\(\sum_{i = l}^rw*a_{i} = w * \sum_{i = l}^{r}\)
所以这题就是一个区间乘区间求和的模板题了,注意开\(long\;long\)就好
#include
#include
using namespace std;
typedef long long ll;
const int maxn = 100100;
const int mod = 1e9 + 7;
struct matrix{//矩阵模板
ll val[4][4];
int x,y;
matrix operator * (const matrix &rhs)const{
matrix ret;
ret.x = x;
ret.y = rhs.y;
for(int i = 1;i <= ret.x;i++)
for(int j = 1;j <= ret.y;j++){
ret.val[i][j] = 0;
for(int k = 1;k <= y;k++)
ret.val[i][j] += val[i][k] * rhs.val[k][j],ret.val[i][j] %= mod;
}
return ret;
}
matrix operator + (const matrix &rhs)const{
matrix ret;
ret.x = rhs.x;
ret.y = rhs.y;
for(int i = 1;i <= ret.x;i++)
for(int j = 1;j <= ret.y;j++)
ret.val[i][j] = (val[i][j] + rhs.val[i][j]) % mod;
return ret;
}
inline void print(){
for(int i = 1;i <= x;i++){
for(int j = 1;j <= y;j++)
printf("%lld ",val[i][j]);
printf("\n");
}
printf("\n");
}
}orgin,w,unit;
inline void init(){//初始化,orgin上文已经提到,w为转移矩阵,unit为单位矩阵
orgin.x = 1;
orgin.y = 2;
orgin.val[1][1] = 0;
orgin.val[1][2] = 1;
w.x = w.y = 2;
w.val[1][1] = 1;
w.val[1][2] = 1;
w.val[2][1] = 1;
w.val[2][2] = 0;
unit.x = unit.y = 2;
unit.val[1][1] = 1;
unit.val[2][2] = 1;
}
namespace ST{//线段树模板
struct Node{
int l,r;
matrix mark,val;
}tree[maxn << 2];
#define lson (root << 1)
#define rson (root << 1 | 1)
inline void maintain(int root){
tree[root].val = tree[lson].val + tree[rson].val;
}
inline void pushdown(int root){
tree[lson].val = tree[lson].val * tree[root].mark;
tree[lson].mark = tree[lson].mark * tree[root].mark;
tree[rson].val = tree[rson].val * tree[root].mark;
tree[rson].mark = tree[rson].mark * tree[root].mark;
tree[root].mark = unit;
}
inline void build(int a,int b,int root = 1){
tree[root].l = a;
tree[root].r = b;
tree[root].mark = unit;
if(a == b){
tree[root].val = orgin;
return;
}
int mid = (tree[root].l + tree[root].r) >> 1;
build(a,mid,lson);
build(mid + 1,b,rson);
maintain(root);
}
inline ll query(int a,int b,int root = 1){
if(a <= tree[root].l && b >= tree[root].r)return tree[root].val.val[1][1];
pushdown(root);
ll ret = 0;
int mid = (tree[root].l + tree[root].r) >> 1;
if(a <= mid)ret += query(a,b,lson),ret %= mod;
if(b >= mid + 1)ret += query(a,b,rson),ret %= mod;
return ret;
}
inline void modify(int a,int b,const matrix val,int root = 1){
if(a <= tree[root].l && b >= tree[root].r){
tree[root].val = tree[root].val * val;
tree[root].mark = tree[root].mark * val;
return;
}
pushdown(root);
int mid = (tree[root].l + tree[root].r) >> 1;
if(a <= mid)modify(a,b,val,lson);
if(b >= mid + 1)modify(a,b,val,rson);
maintain(root);
}
#undef lson
#undef rson
}
inline matrix power(const matrix &a,int b){//快速幂
matrix ret = unit,base = a;
while(b){
if(b & 1)ret = ret * base;
base = base * base;
b >>= 1;
}
return ret;
}
int n,m;
int main(){
#ifdef LOCAL
freopen("fafa.in","r",stdin);
#endif
init();
scanf("%d %d",&n,&m);
ST::build(1,n);
for(int x,i = 1;i <= n;i++)//实在没想到啥更好的初始化办法了,我还是tcl
scanf("%d",&x),ST::modify(i,i,power(w,x));
for(int i = 1;i <= m;i++){
int opt,l,r,x;
scanf("%d %d %d",&opt,&l,&r);
if(opt == 1)scanf("%d",&x),ST::modify(l,r,power(w,x));
else printf("%lld\n",ST::query(l,r));
}
return 0;
}