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HDU4893--Wow! Such Sequence! （线段树延迟标记）

Wow! Such Sequence!

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3354 Accepted Submission(s): 966

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 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| < 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

Author

Fudan University

Source

2014 Multi-University Training Contest 3

题意：

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. 就这三句话

  1 #include <cmath>
  2 #include <cstdio>
  3 #include <cstdlib>
  4 #include <cstring>
  5 #include <algorithm>
  6 using namespace std;
  7 typedef long long ll;
  8 const int maxn = 1e5 + 100;
  9 ll fab[100];
 10 int lazy[maxn<<2];                   //lazy[pos]为1 表示该区间的数为Fibonacci number。
 11 ll sum1[maxn<<2],sum2[maxn<<2];      // sum1为原数组的和，sum2为变为Fibonacci number之后的和，维护这两个数组
 12 void pre_solve()                     //预处理数前90个斐波那契数
 13 {
 14     fab[0] = fab[1] = 1;
 15     for (int i = 2; i <= 90; i++)
 16         fab[i] = fab[i-1] + fab[i-2];
 17 }
 18 ll find_fab(ll x)                          //找到距离x最近的斐波那契数
 19 {
 20     ll ans = fab[0],delta= abs(fab[0] - x);
 21     for (int i = 0; i <= 90; i++)
 22     {
 23         if (delta > abs(x - fab[i]))
 24         {
 25             delta = abs (x - fab[i]);
 26             ans = fab[i];
 27         }
 28     }
 29     return ans;
 30 }
 31 void push_down(int pos)
 32 {
 33     if (lazy[pos])
 34     {
 35         lazy[pos<<1] = lazy[pos<<1|1] = lazy[pos];
 36         lazy[pos] = 0;
 37         sum1[pos<<1] = sum2[pos<<1];
 38         sum1[pos<<1|1] = sum2[pos<<1|1];
 39     }
 40 }
 41 void push_up(int pos)
 42 {
 43     sum1[pos] = sum1[pos<<1] + sum1[pos<<1|1];
 44     sum2[pos] = sum2[pos<<1] + sum2[pos<<1|1];
 45 }
 46 void build(int l,int r,int pos)
 47 {
 48     sum1[pos] = lazy[pos] = 0;
 49     if (l == r)
 50     {
 51         sum2[pos] = 1;                          //初始状态 所有数都为0，距离0最近的斐波那契数是1
 52         return;
 53     }
 54     int mid = (l + r) >> 1;
 55     build(l,mid,pos<<1);
 56     build(mid+1,r,pos<<1|1);
 57     push_up(pos);
 58 }
 59 void update_add(int l,int r,int pos,int x,int val)
 60 {
 61     if (l == r)
 62     {
 63         if (lazy[pos])
 64             sum1[pos] = sum2[pos] + val;            //如果该位置的数字为斐波那契数，那么在此基础上加val
 65         else
 66             sum1[pos] += val;                      //不是斐波那契数 直接加val
 67         sum2[pos] = find_fab(sum1[pos]);            //sum1[pos] 改变，相应的sum2也要改变
 68         lazy[pos] = 0;
 69         return;
 70     }
 71     push_down(pos);
 72     int mid = (l + r) >> 1;
 73     if (x <= mid)
 74         update_add(l,mid,pos<<1,x,val);
 75     else
 76         update_add(mid+1,r,pos<<1|1,x,val);
 77     push_up(pos);
 78 }
 79 void update_fab(int l,int r,int pos,int ua,int ub)
 80 {
 81     if (ua <= l && ub >= r)
 82     {
 83         sum1[pos] = sum2[pos];
 84         lazy[pos] = 1;
 85         return;
 86     }
 87     push_down(pos);
 88     int mid = (l + r) >> 1;
 89     if (ua <= mid)
 90         update_fab(l,mid,pos<<1,ua,ub);
 91     if (ub > mid)
 92         update_fab(mid+1,r,pos<<1|1,ua,ub);
 93     push_up(pos);
 94 }
 95 ll query(int l,int r,int pos,int ua,int ub)
 96 {
 97     if (ua <= l && ub >= r )
 98         return sum1[pos];
 99     push_down(pos);
100     int mid = (l + r) >> 1;
101     ll ans = 0;
102     if (ua <= mid)
103         ans += query(l,mid,pos<<1,ua,ub);
104     if (ub > mid)
105         ans += query(mid+1,r,pos<<1|1,ua,ub);
106     return ans;
107 }
108 int main(void)
109 {
110     #ifndef ONLINE_JUDGE
111         freopen("in.txt","r",stdin);
112         freopen("out.txt","w",stdout);
113     #endif
114     pre_solve();
115     int n,m;
116     while (~scanf ("%d%d",&n,&m))
117     {
118         build(1,n,1);
119         for (int i = 0; i < m; i++)
120         {
121             int op,x,y;
122             scanf ("%d%d%d",&op,&x,&y);
123             if (op == 1)
124                 update_add(1,n,1,x,y);
125             if (op == 3)
126                 update_fab(1,n,1,x,y);
127             if (op == 2)
128                 printf("%I64d
",query(1,n,1,x,y));
129         }
130     }
131     return 0;
132 }

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原文地址：https://www.cnblogs.com/oneshot/p/4089229.html

HDU4893--Wow! Such Sequence! （线段树 延迟标记）

Wow! Such Sequence!

HDU4893--Wow! Such Sequence! （线段树延迟标记）