题意是这样的，给定一个n个元素的数组，初始值为0，3种操作：
1 k d将第k个数增加d；
2 l r 询问区间l...r范围内数之和；
3 l r 表示将区间l...r内的数变成离他最近的斐波那契数，要求尽量小。
线段树操作题目，其中对于第三种操作用一个懒惰标记一下，表示l...r内的数是不是已经变成斐波那契数，如果是的话，求和就是其相应数的斐波那契数之和。
代码：
1 //Template updates date: 20140718
2 #include <bits/stdc++.h>
3 #define esp 1e-6
4 #define inf 0x3f3f3f3f
5 #define pi acos(-1.0)
6 #define pb push_back
7 #define lson l, m, rt<<1
8 #define rson m+1, r, rt<<1|1
9 #define lowbit(x) (x&(-x))
10 #define mp(a, b) make_pair((a), (b))
11 #define bit(k) (1<<(k))
12 #define iin freopen("pow.in", "r", stdin);
13 #define oout freopen("pow.out", "w", stdout);
14 #define in freopen("solve_in.txt", "r", stdin);
15 #define out freopen("solve_out.txt", "w", stdout);
16 #define bug puts("********))))))");
17 #define Inout iin oout
18 #define inout in out
19
20 #define SET(a, v) memset(a, (v), sizeof(a))
21 #define SORT(a) sort((a).begin(), (a).end())
22 #define REV(a) reverse((a).begin(), (a).end())
23 #define READ(a, n) {REP(i, n) cin>>(a)[i];}
24 #define REP(i, n) for(int i = 0; i < (n); i++)
25 #define VREP(i, n, base) for(int i = (n); i >= (base); i--)
26 #define Rep(i, base, n) for(int i = (base); i < (n); i++)
27 #define REPS(s, i) for(int i = 0; (s)[i]; i++)
28 #define pf(x) ((x)*(x))
29 #define mod(n) ((n))
30 #define Log(a, b) (log((double)b)/log((double)a))
31 #define Srand() srand((int)time(0))
32 #define random(number) (rand()%number)
33 #define random_range(a, b) (int)(((double)rand()/RAND_MAX)*(b-a) + a)
34
35 using namespace std;
36 typedef long long LL;
37 typedef unsigned long long ULL;
38 typedef vector<int> VI;
39 typedef pair<int,int> PII;
40 typedef vector<PII> VII;
41 typedef vector<PII, int> VIII;
42 typedef VI:: iterator IT;
43 typedef map<string, int> Mps;
44 typedef map<int, int> Mpi;
45 typedef map<int, PII> Mpii;
46 typedef map<PII, int> Mpiii;
47 const int maxm = 140000 + 1000;
48
49 LL f[92];
50 int n, m;
51 LL sum[maxm<<2], sum1[maxm<<2], cover[maxm<<2];
52
53 void pre() {
54 f[0] = f[1] = 1;
55 Rep(i, 2, 92) {
56 f[i] = f[i-1] + f[i-2];
57 }
58 }
59 void PushDown(int rt) {
60 if(cover[rt]) {
61 cover[rt] = 0;
62 cover[rt<<1] = cover[rt<<1|1] = 1;
63 sum[rt<<1] = sum1[rt<<1];
64 sum[rt<<1|1] = sum1[rt<<1|1];
65 }
66 }
67 void PushUp(int rt) {
68 sum[rt] = sum[rt<<1]+sum[rt<<1|1];
69 sum1[rt] = sum1[rt<<1]+sum1[rt<<1|1];
70 }
71 void build(int l, int r, int rt) {
72 if(l == r) {
73 sum[rt] = 0;
74 sum1[rt] = 1;
75 cover[rt] = 0;
76 return ;
77 }
78 int m = (l+r)>>1;
79 build(lson);
80 build(rson);
81 PushUp(rt);
82 cover[rt] = 0;
83 }
84
85 void update(int l, int r, int rt, int k, int d) {
86 if(l == k && r == k) {
87 sum[rt] += d;
88 int b = upper_bound(f+1, f+92, sum[rt]) - f;
89 if(abs(f[b]-sum[rt]) >= abs(f[b-1]-sum[rt]))
90 sum1[rt] = f[b-1];
91 else
92 sum1[rt] = f[b];
93 return;
94 }
95 PushDown(rt);
96 int m = (l+r)>>1;
97 if(k <= m)
98 update(lson, k, d);
99 else update(rson, k, d);
100 PushUp(rt);
101 }
102 void update1(int L, int R, int l, int r, int rt) {
103 if(L <=l && R >= r) {
104 cover[rt] = 1;
105 sum[rt] = sum1[rt];
106 return;
107 }
108 PushDown(rt);
109 int m = (l+r)>>1;
110 if(L <= m)
111 update1(L, R, lson);
112 if(R > m)
113 update1(L, R, rson);
114 PushUp(rt);
115 }
116 LL query(int L, int R, int l, int r, int rt) {
117 if(L <= l && R >= r) {
118 return sum[rt];
119 }
120 int m = (l+r)>>1;
121 PushDown(rt);
122 LL ans = 0;
123 if(L <= m)
124 ans += query(L, R, lson);
125 if( R > m)
126 ans += query(L, R, rson);
127 return ans;
128 }
129 int main() {
130
131 pre();
132 while(scanf("%d%d", &n, &m) == 2) {
133 build(1, n, 1);
134 REP(i, m) {
135 int u, l, r;
136 scanf("%d%d%d", &u, &l, &r);
137 if(u == 2) {
138 printf("%I64d\n", query(l, r, 1, n, 1));
139 } else if(u == 1) {
140 update(1, n, 1, l, r);
141 } else {
142 update1(l, r, 1, n, 1);
143 }
144 }
145 }
146 return 0;
147 }
HDOJ 4893 Wow! Such Sequence!,布布扣,bubuko.com
时间： 2024-10-14 11:30:10