HDU4893线段树单点跟新+二分+laz标记

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

Wow! Such Sequence!

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1932 Accepted Submission(s): 591

Problem 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 F ₀ = 1,F ₁ = 1,Fibonacci number Fn is defined as F _n = F _{n - 1} + F _{n - 2} for n ≥ 2.

Nearest Fibonacci number of number x means the smallest Fn where |F _n - 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| < 2 ³¹, 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

#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>
#include <ctype.h>
#include <limits.h>
#include <math.h>
#include <string.h>
#include <string>
#include <algorithm>
#include <iostream>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <vector>
#include <map>
using namespace std;
#define int64 __int64
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
const int maxn = 200000;
struct node{int64 l;int64 r;int64 sum;bool laz;//int64 mid(){// return (l+r)/2;//}
};
struct node tree[maxn<<2];
set<int64> ss;
int64 f[10000];void PushUp(int64 rt){if(tree[rt<<1].laz==1 && tree[rt<<1|1].laz==1){//如果两个子节点都不需要更新，那么父节点也不需要跟新tree[rt].laz = 1;}else{tree[rt].laz = 0;}
}
void getf(){int i;f[0] = 1;f[1] = 1;ss.insert(f[0]);ss.insert(f[1]);for(i=2;i<=73;i++){f[i] = f[i-1]+f[i-2];ss.insert(f[i]);}
}void build(int64 l,int64 r,int64 rt){tree[rt].l = l;tree[rt].r = r;tree[rt].laz = 0;if(tree[rt].l == tree[rt].r){tree[rt].sum = 0;return ;}int64 m = (tree[rt].l+tree[rt].r)/2;build(lson);build(rson);tree[rt].sum = tree[rt<<1].sum + tree[rt<<1|1].sum;
}void update(int64 a,int64 b,int64 rt){if(tree[rt].l == tree[rt].r){tree[rt].sum+=b;tree[rt].laz = 0;return ;}int64 m = (tree[rt].l+tree[rt].r)/2;if(a <= m){update(a,b,rt<<1);}else{update(a,b,rt<<1|1);}tree[rt].sum = tree[rt<<1].sum+tree[rt<<1|1].sum;PushUp(rt);
}void update2(int64 L,int64 R,int64 rt){if(tree[rt].laz == 1){return ;}if(tree[rt].l == tree[rt].r){tree[rt].laz = 1;set<int64>::iterator it1,it2;int64 l,r;it2=it1=ss.lower_bound(tree[rt].sum);//获得》=tree[rt].sum这个数在斐波那契数列中的地址l = *it2;if(it1 != ss.begin()){it1--;}r = *it1;tree[rt].sum = (tree[rt].sum-*it1)>((*it2)-tree[rt].sum)?l:r;return ;}int64 m = (tree[rt].l + tree[rt].r)/2;if(L<=m){update2(L,R,rt<<1);}if(R > m){update2(L,R,rt<<1|1);}tree[rt].sum = tree[rt<<1].sum + tree[rt<<1|1].sum;PushUp(rt);
}int64 query(int64 L,int64 R,int64 rt){if(L<=tree[rt].l && tree[rt].r<=R){return tree[rt].sum;}int64 m = (tree[rt].l+tree[rt].r)/2;int64 ret = 0;if(L <= m){ret+=query(L,R,rt<<1);}if(R > m){ret+=query(L,R,rt<<1|1);}//else//{// ret+=query(L,R,rt<<1);//    ret+=query(L,R,rt<<1|1);//}return ret;
}int main(){int64 n,m;int64 op,a,b;ss.clear();getf();while(~scanf("%I64d%I64d",&n,&m)){build(1,n,1);while(m--){scanf("%I64d",&op);if(op == 2){scanf("%I64d%I64d",&a,&b);printf("%I64d\n",query(a,b,1));}else if(op == 1){scanf("%I64d%I64d",&a,&b);update(a,b,1);}else if(op == 3){scanf("%I64d%I64d",&a,&b);update2(a,b,1);}}}return 0;
}

HDU4893线段树单点跟新+二分+laz标记相关推荐

HDUOJ----1166敌兵布阵（线段树单点更新）
敌兵布阵 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submi ...
poj 2892---Tunnel Warfare（线段树单点更新、区间合并）
题目链接 Description During the War of Resistance Against Japan, tunnel warfare was carried out extensiv ...
HDU - 1166敌兵布阵+HDU-1754 I Hate It （线段树单点更新——累加/最大值）
线段树单点更新,模板题 HDU1166 敌兵布阵 C国的死对头A国这段时间正在进行军事演习,所以C国间谍头子Derek和他手下Tidy又开始忙乎了.A国在海岸线沿直线布置了N个工兵营地,Derek和T ...
【原创】tyvj1038 忠诚计蒜客管家的忠诚线段树(单点更新，区间查询)...
[原创]tyvj1038 忠诚 & 计蒜客管家的忠诚 & 线段树(单点更新,区间查询) 最简单的线段树之一,中文题目,不翻译.... 注释讲的比较少,这已经是最简单的线段树,如果看不 ...
FZU 2297 Number theory【线段树/单点更新/思维】
Given a integers x = 1, you have to apply Q (Q ≤ 100000) operations: Multiply, Divide. Input First l ...
CDOJ 1073 线段树单点更新+区间查询水题
H - 秋实大哥与线段树 Time Limit:1000MS Memory Limit:65535KB 64bit IO Format:%lld & %llu Submit S ...
CodeforcesBeta Round #19 D. Points 离线线段树单点更新离散化
题目链接: http://codeforces.com/contest/19/problem/D 题意: 有三种操作"add x y"往平面上添加(x,y)这个点,"re ...
线段树——单点更新（二）
HDU 4217 Data Structure? http://acm.hdu.edu.cn/showproblem.php?pid=4217 CZ做的一道题目,我帮忙看了看. 题意:给定N个数(1- ...
Ocean的礼物（线段树单点修改）
题目链接:http://oj.ismdeep.com/contest/Problem?id=1284&pid=0 A: Ocean的礼物 Time Limit: 5 s Memory ...
P3834 【模板】可持久化线段树 2（整体二分做法）
P3834 [模板]可持久化线段树 2(主席树) 我们详细讲讲这个整体二分如何求区间第k小我们都知道二分可以求出区间里某个想要的值,如果有很多询问,我们对每个询问都进行二分,复杂度就是O(QNlog ...

HDU4893线段树单点跟新+二分+laz标记

Wow! Such Sequence!

HDU4893线段树单点跟新+二分+laz标记相关推荐

最新文章

热门文章