POJ 3468 A Simple Problem with Integers（线段树）

Posted 2021-01-21 ehanla

题目链接：A Simple Problem with Integers

题意：N个数字，M次操作；操作分为两种，第一种将$[L,R]$区间内的每个数都加上C，第二种为求$[L,R]$区间内的数字和。

题解：线段树基本操作，区间修改和区间求和

 1 #include <cstdio>
 2 #define LC(a) ((a<<1))
 3 #define RC(a) ((a<<1)+1)
 4 #define MID(a,b) ((a+b)>>1)
 5 using namespace std;
 6 
 7 typedef long long ll;
 8 const int N=5e5*4;
 9 ll ans=0;
10 
11 struct node{
12     ll l,r;
13     ll sum,add;
14 }tree[N];
15 
16 void pushup(ll p){
17     tree[p].sum=tree[LC(p)].sum+tree[RC(p)].sum;
18 }
19 
20 void pushdown(ll p){
21     tree[LC(p)].add+=tree[p].add;
22     tree[RC(p)].add+=tree[p].add;
23     tree[LC(p)].sum+=tree[p].add*(tree[LC(p)].r-tree[LC(p)].l+1);
24     tree[RC(p)].sum+=tree[p].add*(tree[RC(p)].r-tree[RC(p)].l+1);
25     tree[p].add=0;
26 }
27 
28 void build(ll p,ll l,ll r){
29     tree[p].l=l;
30     tree[p].r=r;
31     tree[p].sum=tree[p].add=0;
32     if(l==r){
33         scanf("%lld",&tree[p].sum);
34         return;
35     }
36     build(LC(p),l,MID(l,r));
37     build(RC(p),MID(l,r)+1,r);
38     pushup(p);
39 }
40 
41 void updata(ll p,ll l,ll r,ll num){
42     if(r<tree[p].l||l>tree[p].r) return;
43     if(l<=tree[p].l&&r>=tree[p].r){
44         tree[p].add+=num;
45         tree[p].sum+=num*(tree[p].r-tree[p].l+1);
46         return;
47     }
48     if(tree[p].add) pushdown(p);
49     updata(LC(p),l,r,num);
50     updata(RC(p),l,r,num);
51     pushup(p);
52 }
53 
54 void query(ll p,ll l, ll r){
55     if(r<tree[p].l||l>tree[p].r) return;
56     if(l<=tree[p].l&&r>=tree[p].r){
57         ans+=tree[p].sum;
58         return;
59     }
60     if(tree[p].add) pushdown(p);
61     query(LC(p),l,r);
62     query(RC(p),l,r);
63 }
64 
65 int main(){
66     ll n,q;
67     char op[5];
68     scanf("%lld%lld",&n,&q);
69     build(1,1,n);
70     while(q--){
71         scanf("%s",op);
72         if(op[0]==‘C‘){
73             ll l,r,add;
74             scanf("%lld%lld%lld",&l,&r,&add);
75             updata(1,l,r,add);
76         }
77         else{
78             ans=0;
79             ll l,r;
80             scanf("%lld%lld",&l,&r);
81             query(1,l,r);
82             printf("%lld
",ans);
83         }
84     }
85     return 0;
86 }

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