CF438 The Child and Sequence

题意:

给定一个长度为n的非负整数序列a,你需要支持以下操作:
1)给定l,r,输出a[l] + a[l+1] + ... + a[r]

2)给定l,r,x, 将a[l]、a[l+1]、....、a[r]x取模

3)给定k,y,将a[k]修改为y

n, m <= 100000,a[i], x, y <= 109


对于操作(1)(3)非常简单,线段树基本操作

问题是操作(2),显然的是我们不能对区间和取模,这样就很难受

但是我们可以想到,一个数若是比模数小,就不需要取模,而一个数w有效取模次数最多为log(w)

同时单个数被有效取模的一次只会花费O(logn)

因此每次修改至多使复杂度增加O(lognlogw)

这样我们对于区间l, r暴力对每个能取模的数取模即可

最后时间复杂度为O(mlognlogw)

 

 1 #include<bits/stdc++.h>
 2 #define ll long long
 3 #define uint unsigned int
 4 #define ull unsigned long long
 5 using namespace std;
 6 const int maxn = 100010;
 7 struct shiki {
 8     ll maxx, sum;
 9 }tree[maxn << 2];
10 int n, m;
11 ll a[maxn];
12 
13 inline ll read() {
14     ll x = 0, y = 1;
15     char ch = getchar();
16     while(!isdigit(ch)) {
17         if(ch == '-') y = -1;
18         ch = getchar();
19     }
20     while(isdigit(ch)) {
21         x = (x << 1) + (x << 3) + ch - '0';
22         ch = getchar();
23     }
24     return x * y;
25 }
26 
27 inline void maintain(int pos) {
28     int ls = pos << 1, rs = pos << 1 | 1;
29     tree[pos].maxx = max(tree[ls].maxx, tree[rs].maxx);
30     tree[pos].sum = tree[ls].sum + tree[rs].sum;
31 }
32 
33 void build(int pos, int l, int r) {
34     if(l == r) {
35         tree[pos].maxx = tree[pos].sum = a[l];
36         return;
37     }
38     int mid = l + r >> 1;
39     build(pos << 1, l, mid);
40     build(pos << 1 | 1, mid + 1, r);
41     maintain(pos);
42 }
43 
44 void get_mod(int pos, int L, int R, int l, int r, ll mod) {
45     if(l > R || r < L) return;
46     if(tree[pos].maxx < mod) return;
47     if(l == r) {
48         tree[pos].sum %= mod;
49         tree[pos].maxx %= mod;
50         return;
51     }
52     int mid = l + r >> 1;
53     get_mod(pos << 1, L, R, l, mid, mod);
54     get_mod(pos << 1 | 1, L, R, mid + 1, r, mod);
55     maintain(pos);
56 }
57 
58 void update(int pos, int aim, int l, int r, ll val) {
59     if(l == r && l == aim) {
60         tree[pos].maxx = tree[pos].sum = val;
61         return;
62     }
63     int mid = l + r >> 1;
64     if(aim <= mid) update(pos << 1, aim, l, mid, val);
65     else update(pos << 1 | 1, aim, mid + 1, r, val);
66     maintain(pos);
67 }
68 
69 ll query_sum(int pos, int L, int R, int l, int r) {
70     if(l > R || r < L) return 0;
71     if(l >= L & r <= R) return tree[pos].sum;
72     int mid = l + r >> 1;
73     return query_sum(pos << 1, L, R, l, mid) + query_sum(pos << 1 | 1, L, R, mid + 1, r); 
74 }
75 
76 int main() {
77     n = read(), m = read();
78     for(int i = 1; i <= n; ++i) a[i] = read();
79     build(1, 1, n);
80     for(int i = 1; i <= m; ++i) {
81         int opt = read(), x = read(), y = read();
82         if(opt == 1) printf("%I64d\n", query_sum(1, x, y, 1, n));
83         if(opt == 2) {
84             ll p = read();
85             get_mod(1, x, y, 1, n, p); 
86         }
87         if(opt == 3) update(1, x, 1, n, y);
88     } 
89     return 0;
90 }

 

posted @ 2019-02-17 14:25 YuWenjue 阅读(...) 评论(...) 编辑 收藏