线段树

这是洛谷线段树模板题绿标题，线段树好像没什么好总结的，主要看脑子

 1 #include<iostream>
 2 using namespace std;
 3 #define int long long
 4 const int N = 1e5 + 10, inf = 0x3f3f3f3f3f3f3f3f;
 5 int n, q, m;
 6 int f[4 * N], a[N], lazy1[4 * N], lazy2[4 * N];
 7 //lazy标记更新操作
 8 void tag(int k, int l, int mid, int r) {
 9     f[2 * k] = (lazy2[k] * f[2 * k] % m + ((mid - l + 1) * lazy1[k]) % m) % m;
10     f[2 * k + 1] = (lazy2[k] * f[2 * k + 1] % m + ((r - mid) * lazy1[k]) % m) % m;
11     lazy2[2 * k] = (lazy2[2 * k] * lazy2[k]) % m;
12     lazy2[2 * k + 1] = (lazy2[2 * k + 1] * lazy2[k]) % m;
13     lazy1[2 * k] = ((lazy1[2 * k] * lazy2[k]) % m + lazy1[k]) % m;
14     lazy1[2 * k + 1] = ((lazy1[2 * k + 1] * lazy2[k]) % m + lazy1[k]) % m;
15     lazy1[k] = 0; lazy2[k] = 1;
16 }
17 //建树
18 void build(int k, int l, int r) {
19     lazy2[k] = 1; lazy1[k] = 0;
20     if (l == r) {
21         f[k] = a[l] % m;
22         return;
23     }
24     int mid = (l + r) >> 1;
25     build(2 * k, l, mid);
26     build(2 * k + 1, mid + 1, r);
27     f[k] = (f[2 * k] + f[2 * k + 1]) % m;
28 }
29 //区间加法操作
30 void add(int k, int l, int r, int x, int y, int val) {
31     if (l >= x && r <= y) {
32         f[k] = (f[k] + ((r - l + 1) * val) % m) % m;
33         lazy1[k] = (lazy1[k] + val) % m;
34         return;
35     }
36     int mid = (l + r) >> 1;
37     tag(k, l, mid, r);
38     if (x <= mid)add(2 * k, l, mid, x, y, val);
39     if (y > mid)add(2 * k + 1, mid + 1, r, x, y, val);
40     f[k] = (f[2 * k] + f[2 * k + 1]) % m;
41 }
42 //区间乘法操作
43 void mul(int k, int l, int r, int x, int y, int val) {
44     if (l >= x && r <= y) {
45         f[k] = (f[k] * val) % m;
46         lazy2[k] = (lazy2[k] * val) % m;
47         lazy1[k] = (lazy1[k] * val) % m;
48         return;
49     }
50     int mid = (l + r) >> 1;
51     tag(k, l, mid, r);
52     if (x <= mid)mul(2 * k, l, mid, x, y, val);
53     if (y > mid)mul(2 * k + 1, mid + 1, r, x, y, val);
54     f[k] = (f[2 * k] + f[2 * k + 1]) % m;
55 }
56 //查询区间总和操作
57 int check(int k, int l, int r, int x, int y) {
58     if (l >= x && r <= y) return f[k];
59     int mid = (l + r) >> 1;
60     tag(k, l, mid, r);
61     int ans1 = 0, ans2 = 0;
62     if (x <= mid)ans1 = check(2 * k, l, mid, x, y);
63     if (y > mid)ans2 = check(2 * k + 1, mid + 1, r, x, y);
64     return (ans1 % m + ans2 % m) % m;
65 }
66 void solve() {
67     cin >> n >> q >> m;
68     for (int i = 1; i <= n; i++)cin >> a[i];
69     build(1, 1, n);
70     while (q--) {
71         int op;
72         cin >> op;
73         if (op == 2) {
74             int x, y, k;
75             cin >> x >> y >> k;
76             add(1, 1, n, x, y, k);
77         }
78         else if (op == 1) {
79             int x, y, k;
80             cin >> x >> y >> k;
81             mul(1, 1, n, x, y, k);
82         }
83         else if (op == 3) {
84             int x, y;
85             cin >> x >> y;
86             int res = check(1, 1, n, x, y);
87             cout << res % m << "\n";
88         }
89     }
90 }
91 signed main() {
92     ios::sync_with_stdio(false);
93     cin.tie(0);
94     cout.tie(0);
95     int t = 1;
96     //cin >> t;
97     while (t--)solve();
98     return 0;
99 }

那么多行，wa后调试真的会疯。。。