[模板]线段树1

https://www.luogu.org/problemnew/show/P3372

 1 #include <bits/stdc++.h>
 2 #define lowbit(a) (a&-a)
 3
 4 using namespace std;
 5 typedef long long LL;
 6
 7 const LL N = 100010;
 8 LL n, m, q, c[N], c2[N], opt, x, y, k;
 9 inline LL Sum(LL *a, LL x){
10     LL ans = 0;
11     for(LL i = x; i; i -= lowbit(i)) ans += a[i];
12     return ans;
13 }
14
15 inline void add(LL *a, LL x, LL q){
16     for(LL i = x; i <= n; i += lowbit(i)) a[i] += q;
17 }
18
19 inline void read(LL &x){
20     register LL c = getchar(), f = 1;x = 0;
21     while(!isdigit(c)) {if (c == ‘-‘) f = -1; c = getchar();}
22     while(isdigit(c)) x = (x<<3)+(x<<1)+(c^48), c = getchar();
23     x *= f;
24 }
25
26 int main(void){
27     LL now, tmp = 0;
28     read(n); read(m);
29     for(register LL i = 1; i <= n; ++i){
30         read(now);
31         add(c, i, now-tmp);
32         add(c2, i, (i-1)*(now-tmp));
33         tmp = now;
34     }
35     while(m--){
36         LL opt; read(opt);
37         if (opt == 1){
38             read(x), read(y), read(k);
39             add(c, x, k), add(c, y+1, -k), add(c2, x, (x-1)*k), add(c2, y+1, -k*y);
40         }
41         else {
42             read(x), read(y);
43             LL sum1 = (x-1)*Sum(c,x-1)-Sum(c2,x-1),
44                  sum2 = y*Sum(c,y)-Sum(c2,y);
45             printf("%lld\n", sum2-sum1);
46         }
47     }
48     return 0;
49 }

树状数组

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstring>
 5 #include <cmath>
 6 #define lson (o<<1)
 7 #define rson (o<<1|1)
 8 #define midd int mid = (l+r)>>1
 9 using namespace std;
10 const int N = 100010;
11
12 int a[N], n, m;
13
14 inline int read() {
15     register int n, ch;
16     n = 0, ch = getchar();
17     while (!isdigit(ch)) ch = getchar();
18     while (isdigit(ch)) n = (n << 3) + (n << 1) + ch - ‘0‘, ch = getchar();
19     return n;
20 }
21 struct Node{
22     long long sum, add;
23 }t[N<<2];
24 inline void pushup(int o){t[o].sum=t[lson].sum+t[rson].sum;}
25 inline void pushdown(int o, int l, int r){
26     int ls = lson, rs = rson, mid = (l+r)>>1;
27   t[ls].add+=t[o].add;
28   t[rs].add+=t[o].add;
29   t[ls].sum+=(mid-l+1)*t[o].add;
30   t[rs].sum+=(r-mid)*t[o].add;
31   t[o].add=0;
32 }
33 inline void secadd(int o,int l, int r, int ql, int qr, int v){
34     if (ql<=l&&r<=qr){
35         t[o].add+=v;
36         t[o].sum+=(r-l+1)*v;
37         return ;
38     }
39     int mid = (l+r)>>1;
40       if (t[o].add) pushdown(o, l, r);
41       if (ql<=mid) secadd(lson,l,mid,ql,qr,v);
42         if (qr>mid) secadd(rson,mid+1,r,ql,qr,v);
43     pushup(o);
44 }
45 inline long long query(int o,int l, int r, int ql, int qr){
46     if (ql<=l&&r<=qr)
47         return t[o].sum;
48     if (t[o].add) pushdown(o,l,r);int mid = (l+r)>>1;long long ans = 0;
49     if (ql<=mid) ans+=query(lson,l,mid,ql,qr);
50         if (qr>mid) ans += query(rson,mid+1,r,ql,qr);
51         return ans;
52 }
53 inline void build(int o, int l, int r){
54     if (l==r){t[o].sum=a[l];return ;}
55     int mid = (l+r)>>1;
56     build(lson,l,mid), build(rson,mid+1,r);
57     pushup(o);
58 }
59
60 int main(void){
61     n=read(), m=read();
62     for(register int i = 1; i <= n; ++i){
63         a[i] = read();
64     }
65     build(1,1,n);
66     while(m--){
67         register int bs = read(), x = read(), y = read(), k;
68         if (bs == 1){k = read(); secadd(1,1,n,x,y,k);}
69         else printf("%lld\n", query(1,1,n,x,y));
70     }
71     return 0;
72 }

线段树

原文地址：https://www.cnblogs.com/Ycrpro/p/8452481.html