HDU 4893 Wow! Such Sequence! （线段树）

Wow! Such Sequence!

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=4893

Description

Recently, Doge got a funny birthday present from his new friend, Protein Tiger from St. Beeze College. No, not cactuses. It's a mysterious blackbox.

After some research, Doge found that the box is maintaining a sequence an of n numbers internally, initially all numbers are zero, and there are THREE "operations":

1.Add d to the k-th number of the sequence.
2.Query the sum of ai where l ≤ i ≤ r.
3.Change ai to the nearest Fibonacci number, where l ≤ i ≤ r.
4.Play sound "Chee-rio!", a bit useless.

Let F0 = 1,F1 = 1,Fibonacci number Fn is defined as Fn = Fn - 1 + Fn - 2 for n ≥ 2.

Nearest Fibonacci number of number x means the smallest Fn where |Fn - x| is also smallest.

Doge doesn't believe the machine could respond each request in less than 10ms. Help Doge figure out the reason.

Input

Input contains several test cases, please process till EOF.
For each test case, there will be one line containing two integers n, m.
Next m lines, each line indicates a query:

1 k d - "add"
2 l r - "query sum"
3 l r - "change to nearest Fibonacci"

1 ≤ n ≤ 100000, 1 ≤ m ≤ 100000, |d| < 231, all queries will be valid.

Output

For each Type 2 ("query sum") operation, output one line containing an integer represent the answer of this query.

Sample Input

1 1

2 1 1

5 4

1 1 7

1 3 17

3 2 4

2 1 5

Sample Output

0

22

题意：

对一个数组进行若干操作:

将A_k增加x.

查询区间和.

将区间内的数变成离它最近的斐波拉契数.

(注意这个斐波拉契数列是从 1 开始的!!!)

题解：

很容易想到用线段树维护，关键是如何操作3. 直接每个点维护复杂度太高.

如果一个数被操作3更新为了斐波那契数，那么再次更新时将不改变它的值. 利用这一点来剪枝.

对于树中的每个结点维护一个isfib标记, 表示当前结点所表示的区间内是否均为斐波那契数.

每次操作1更新时，将该点的isfib标志置0; (别忘了，WA了一次)

每次操作3更新时，若处理到某区间均为fib数则返回，否则一直向下处理到单点，并对单点进行更新.

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include <vector>
#include <list>
#define LL long long
#define eps 1e-8
#define maxn 101000
#define mod 100000007
#define inf 0x3f3f3f3f
#define mid(a,b) ((a+b)>>1)
#define IN freopen("in.txt","r",stdin);
using namespace std;

LL fib[100];

int n;
struct Tree
{
int left,right;
LL sum;
bool isfib;
}tree[maxn<<2];

void pushup(int i) {
tree[i].sum = tree[i<<1].sum + tree[i<<1|1].sum;
tree[i].isfib = tree[i<<1].isfib && tree[i<<1|1].isfib ? 1 : 0;
}

void build(int i,int left,int right)
{
tree[i].left=left;
tree[i].right=right;

if(left==right){
tree[i].sum = 0;
tree[i].isfib = 0;
return ;
}

int mid=mid(left,right);

build(i<<1,left,mid);
build(i<<1|1,mid+1,right);

pushup(i);
}

void update(int i,int x,LL d)
{
if(tree[i].left == tree[i].right){
tree[i].sum += d;
tree[i].isfib = 0;
return;
}

int mid=mid(tree[i].left,tree[i].right);

if(x<=mid) update(i<<1,x,d);
else update(i<<1|1,x,d);

pushup(i);
}

LL find_fib(LL x) {
int pos = lower_bound(fib, fib+90, x) - fib;
if(!pos) return fib[pos];
if(abs(fib[pos]-x) < abs(fib[pos-1]-x)) return fib[pos];
return fib[pos-1];
}

void update_fib(int i,int left,int right)
{
if(tree[i].isfib) return ;
if(tree[i].left == tree[i].right)
{
tree[i].sum = find_fib(tree[i].sum);
tree[i].isfib = 1;
return ;
}

int mid=mid(tree[i].left,tree[i].right);

if(right<=mid) update_fib(i<<1,left,right);
else if(left>mid) update_fib(i<<1|1,left,right);
else {
update_fib(i<<1,left,mid);
update_fib(i<<1|1,mid+1,right);
}

pushup(i);
}

LL query(int i,int left,int right)
{
if(tree[i].left==left&&tree[i].right==right)
return tree[i].sum;

int mid=mid(tree[i].left,tree[i].right);

if(right<=mid) return query(i<<1,left,right);
else if(left>mid) return query(i<<1|1,left,right);
else return query(i<<1,left,mid)+query(i<<1|1,mid+1,right);
}

void init() {
fib[0] = 1; fib[1] = 1;
for(int i=2; i<90; i++) {
fib[i] = fib[i-1] + fib[i-2];
}
}

int main(int argc, char const *argv[])
{
//IN;

init();

int m;
while(scanf("%d %d", &n,&m) != EOF)
{
build(1, 1, n);

while(m--) {
int op, l, r;
scanf("%d %d %d", &op,&l,&r);
if(op == 1) {
update(1, l, (LL)r);
}
else if(op == 2) {
printf("%lld\n", query(1, l, r));
}
else if(op == 3) {
update_fib(1, l, r);
}
}
}

return 0;
}