HDU4893 Wow! Such Sequence! 线段树

题意：给你一个序列，其中有三种操作

1）位置为K 的数+ D

2）求 l-r 区间和

3）把 l-r 区间里面的所有数都变为理它最近的斐波纳契数

解题思路：这个题的区间更新其实可以在单点更新的时候就得出，为节点维护两个 和，一个是 斐波纳契和 一个是正常和 ，再看这个区间有没有被3覆盖，特判一下就行了。

解题代码：

  1 // File Name: 1007.cpp
  2 // Author: darkdream
  3 // Created Time: 2014年07月29日 星期二 12时49分33秒
  4
  5 #include<vector>
  6 #include<list>
  7 #include<map>
  8 #include<set>
  9 #include<deque>
 10 #include<stack>
 11 #include<bitset>
 12 #include<algorithm>
 13 #include<functional>
 14 #include<numeric>
 15 #include<utility>
 16 #include<sstream>
 17 #include<iostream>
 18 #include<iomanip>
 19 #include<cstdio>
 20 #include<cmath>
 21 #include<cstdlib>
 22 #include<cstring>
 23 #include<ctime>
 24 #define LL long long
 25 using namespace std;
 26 #define maxn 100005
 27 struct node{
 28   int is3 ;
 29   int m , r , l ;
 30   LL sum ,fsum ;
 31 }tree[maxn << 2];
 32 LL f[100];
 33 int L(int x)
 34 {
 35    return 2* x;
 36 }
 37 int R(int x)
 38 {
 39    return 2*x + 1;
 40 }
 41 void pushup(int c)
 42 {
 43    tree[c].sum = tree[L(c)].sum + tree[R(c)].sum ;
 44    tree[c].fsum = tree[L(c)].fsum + tree[R(c)].fsum ;
 45 }
 46 void  build(int c, int l , int r )
 47 {
 48      tree[c].l = l ;
 49      tree[c].r = r;
 50      tree[c].m = (l+r)/2;
 51      tree[c].is3 = 0 ;
 52      if(l ==  r)
 53      {
 54         tree[c].sum = 0 ;
 55         tree[c].fsum = 1 ;
 56         return ;
 57      }
 58      build(L(c),l,tree[c].m);
 59      build(R(c),tree[c].m+1,r);
 60      pushup(c);
 61 }
 62 LL ABS(LL x)
 63 {
 64    if(x <= 0 )
 65     return -x;
 66    else return x;
 67 }
 68 LL find( LL x )
 69 {
 70     int l = 1 , r = 85 ;
 71     while(l <= r )
 72     {
 73        int m = (l + r)/2;
 74        if(f[m] > x)
 75        {
 76            r = m - 1;
 77        }else {
 78            l = m + 1;
 79        }
 80     }
 81 //    printf("%I64d %I64d\n",f[l],f[r]);
 82     if(ABS(f[l]-x) < ABS(f[r]-x))
 83     {
 84        return f[l];
 85     }else return f[r];
 86 }
 87 void pushdown(int c)
 88 {
 89     if(tree[c].is3)
 90     {
 91        tree[L(c)].is3 = 1;
 92        tree[R(c)].is3 = 1;
 93        tree[L(c)].sum = tree[L(c)].fsum;
 94        tree[R(c)].sum = tree[R(c)].fsum;
 95        tree[c].is3 = 0 ;
 96     }
 97 }
 98 void update(int c, int k , int d)
 99 {
100      if(tree[c].l == tree[c].r&& tree[c].l == k )
101      {
102          tree[c].is3  = 0 ;
103          tree[c].sum += d;
104          tree[c].fsum = find(tree[c].sum);
105          return ;
106      }
107      pushdown(c);
108      if(k <= tree[c].m)
109          update(L(c),k,d);
110      else update(R(c),k,d);
111      pushup(c);
112 }
113 LL ans = 0 ;
114 void get(int c , int l , int r)
115 {
116        if(l <= tree[c].l && r >= tree[c].r )
117        {
118          ans += tree[c].sum ;
119          // printf("%I64d\n",ans);
120          return ;
121        }
122      pushdown(c);
123      if(l <= tree[c].m)
124          get(L(c),l,r);
125      if(r > tree[c].m)
126          get(R(c),l,r);
127 }
128 void change(int c, int l , int r)
129 {
130       if(tree[c].is3 == 1)
131           return ;
132       if(l <= tree[c].l && r >= tree[c].r)
133         {
134               tree[c].is3 = 1;
135               tree[c].sum = tree[c].fsum;
136               return ;
137         }
138      if(l <= tree[c].m)
139          change(L(c),l,r);
140      if(r > tree[c].m)
141          change(R(c),l,r);
142      pushup(c);
143 }
144 int main(){
145    int n , m;
146    f[0] = 1 ;
147    f[1] = 1;
148    for(int i = 2;i <= 86;i ++)
149        f[i] = f[i-1] + f[i-2];
150    while(scanf("%d %d",&n,&m) != EOF)
151    {
152        build(1 ,1,n);
153        int a, b, c  ;
154        for(int i = 1;i <= m;i ++)
155        {
156          int a, b, c  ;
157          scanf("%d %d %d",&a,&b,&c);
158          if(a == 1 )
159          {
160             update(1,b,c);
161          }else if(a == 2 ){
162             ans = 0 ;
163             get(1,b,c);
164             printf("%I64d\n",ans);
165          }else{
166             change(1,b,c);
167          }
168        }
169    }
170 return 0;
171 }

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