HDU 4893 Wow! Such Sequence! 解题报告（线段树）

Wow! Such Sequence!

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 850    Accepted Submission(s): 251


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 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

 

Author

Fudan University
 

Source

2014 Multi-University Training Contest 3
 
    解题报告：裸裸的线段树，不过很久没接触线段树了，略生疏。代码如下：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <iomanip>
using namespace std;
#define ff(i, n) for(int i=0;i<(n);i++)
#define fff(i, n, m) for(int i=(n);i<=(m);i++)
#define dff(i, n, m) for(int i=(n);i>=(m);i--)
#define bit(n) (1LL<<(n))
typedef long long LL;
typedef unsigned long long ULL;
void work();
int main()
{
#ifdef ACM
freopen("in.txt", "r", stdin);
#endif // ACM
work();
}

/***************************************************/

#define lson l, m, pos<<1
#define rson m+1, r, pos<<1|1

const int maxn = 111111;
LL sum[maxn<<2];
LL fsum[maxn<<2];
bool fib[maxn<<2];

LL fibonacci[100];
int tot = 80;

void updateFather(int pos)
{
sum[pos] = (fib[pos<<1]?fsum[pos<<1]:sum[pos<<1]) + (fib[pos<<1|1]?fsum[pos<<1|1]:sum[pos<<1|1]);
fsum[pos] = fsum[pos<<1] + fsum[pos<<1|1];
}

void build(int l, int r, int pos)
{
fib[pos] = false;

if(l == r)
{
sum[pos] = 0;
fsum[pos] = 1;
fib[pos] = false;
return;
}

int m = (l+r)/2;
build(lson);
build(rson);
updateFather(pos);
}

void updateSon(int pos)
{
if(fib[pos])
{
fib[pos<<1] = fib[pos<<1|1] = true;
fib[pos] = false;
}
}

void updatePoint(int p, int v, int l, int r, int pos)
{
if(l == r)
{
if(fib[pos]) sum[pos] = fsum[pos];
sum[pos] += v;

fib[pos] = false;

int pp = lower_bound(fibonacci+1, fibonacci+tot, sum[pos]) - fibonacci;
if(sum[pos]-fibonacci[pp-1] <= fibonacci[pp]-sum[pos])
pp--;
fsum[pos] = fibonacci[pp];

return;
}

updateSon(pos);

int m = (l+r)/2;
if(p <= m)
updatePoint(p, v, lson);
else
updatePoint(p, v, rson);

updateFather(pos);
}

void updateLine(int L, int R, int l, int r, int pos)
{
if(fib[pos]) return;

if(L<=l && r<=R)
{
fib[pos] = true;
return;
}

int m = (l+r)/2;
if(L <= m)
updateLine(L, R, lson);
if(m < R)
updateLine(L, R, rson);

updateFather(pos);
}

LL query(int L, int R, int l, int r, int pos)
{
if(L<=l && r<=R)
return fib[pos]?fsum[pos]:sum[pos];

updateSon(pos);
updateFather(pos);

int m = (l+r)/2;
return (L<=m ? query(L, R, lson) : 0) + (m<R ? query(L, R, rson) : 0);
}

void init()
{
fibonacci[0] = fibonacci[1] = 1;

fff(i, 2, tot)
fibonacci[i] = fibonacci[i-1] + fibonacci[i-2];
}

void work()
{
init();

int n, m;
while(scanf("%d%d", &n, &m) == 2)
{
build(1, n, 1);

while(m--)
{
int op, l, r;
scanf("%d%d%d", &op, &l, &r);

if(op == 1)
{
updatePoint(l, r, 1, n, 1);
}
else if(op == 2)
{
cout << query(l, r, 1, n, 1) << endl;
}
else if(op == 3)
{
updateLine(l, r, 1, n, 1);
}
}
}
}