LYOI2016 Summer 一次函数 (线段树)

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 typedef long long ll;
 5 const ll mod = 1e9+7;
 6 const int maxn = 210000;
 7 int kk[maxn],bb[maxn];
 8 int n,m;
 9 char op[5];
10 struct node
11 {
12     int l,r;
13     ll k,b;
14 }tree[maxn<<2];
15 node Union (node x,node y)
16 {
17     node tmp;
18     tmp.l=x.l,tmp.r=y.r;
19     tmp.k=(x.k*y.k)%mod;
20     tmp.b=((y.k*x.b)%mod+y.b)%mod;
21     return tmp;
22 }
23 void pushup(int now)
24 {
25     tree[now] = Union(tree[now<<1],tree[now<<1|1]);
26 }
27 void build (int now,int L,int R)
28 {
29     if (L==R){
30         tree[now].l=L,tree[now].r=L;
31         tree[now].k=kk[L];tree[now].b=bb[L];
32         return ;
33     }
34     int mid = (L+R)>>1;
35     build(now<<1,L,mid);
36     build(now<<1|1,mid+1,R);
37     pushup(now);
38 }
39 void update (int now,int pos,int kkk,int bbb)
40 {
41     if (tree[now].l==tree[now].r){
42         tree[now].k=kkk,tree[now].b=bbb;
43         return;
44     }
45     int mid = (tree[now].l+tree[now].r)/2;
46     if (pos<=mid)
47         update(now<<1,pos,kkk,bbb);
48     else
49         update(now<<1|1,pos,kkk,bbb);
50     pushup(now);
51 }
52 node query (int now,int L,int R)
53 {
54     if (tree[now].l==L&&tree[now].r==R)
55         return tree[now];
56     int mid = (tree[now].l+tree[now].r)/2;
57     if (R<=mid)
58         return query(now<<1,L,R);
59     else if (L>mid)
60         return query(now<<1|1,L,R);
61     else
62         return Union(query(now<<1,L,mid),query(now<<1|1,mid+1,R));
63 }
64 int main()
65 {
66     //freopen("de.txt","r",stdin);
67     scanf("%d%d",&n,&m);
68     for (int i=1;i<=n;++i)
69         scanf("%d%d",&kk[i],&bb[i]);
70     build(1,1,n);
71     for (int i=0;i<m;++i){
72         scanf("%s",op);
73         if (op[0]==‘M‘){
74             int x,y,z;
75             scanf("%d%d%d",&x,&y,&z);
76             update(1,x,y,z);
77         }
78         else{
79             int lll,rrr,xx;
80             scanf("%d%d%d",&lll,&rrr,&xx);
81             node anss=query(1,lll,rrr);
82             ll ans = ((anss.k*xx)%mod+anss.b)%mod;
83             printf("%lld\n",ans);
84         }
85     }
86     return 0;
87 }