第一题 线段树
树状数组存差b[i] = a[i]-a[i-1]，反正是单点查询，我为什么没想到。。。很傻的用线段树
#include<bits/stdc++.h> using namespace std; #define maxn 100005 #define ll long long int n, m, a[maxn]; struct SegmentTree{ struct node{ ll sum; int lazy; }; node Tree[maxn << 2]; #define ls l, m, v << 1 #define rs m+1, r, v << 1 | 1 void updata(int v){ Tree[v].sum = Tree[v << 1].sum + Tree[v << 1 | 1].sum; } void push_down(int l, int r, int v){ int m = (l + r) >> 1; Tree[v << 1].sum += Tree[v].lazy * (m - l + 1) * 1LL; Tree[v << 1].lazy += Tree[v].lazy; Tree[v << 1 | 1].sum += Tree[v].lazy * (r - m) * 1LL; Tree[v << 1 | 1].lazy += Tree[v].lazy; Tree[v].lazy = 0; } void build(int l = 1, int r = n, int v = 1){ if(l == r) Tree[v].sum = a[l], Tree[v].lazy = 0; else { int m = (l + r) >> 1; build(ls); build(rs); updata(v); } } void modify(int x, int L, int R, int l = 1, int r = n, int v = 1){ if(l >= L && r <= R){ Tree[v].sum += x * (r - l + 1) * 1LL; Tree[v].lazy += x; } else { if(Tree[v].lazy) push_down(l, r, v); int m = (l + r) >> 1; if(L <= m)modify(x, L, R, ls); if(R > m)modify(x, L, R, rs); updata(v); } } ll query(int x,int l = 1, int r = n, int v = 1){ if(l == r)return Tree[v].sum; else { if(Tree[v].lazy) push_down(l, r, v); int m = (l + r) >> 1; if(x <= m) return query(x, ls); if(x > m)return query(x, rs); } } }; SegmentTree Tr; int main(){ freopen("bit.in","r",stdin); freopen("bit.out","w",stdout); cin>>n; for(int i = 1; i <= n; i++)scanf("%d", a + i); Tr.build(); cin>>m; for(int i = 1; i <= m; i++){ string opt; cin>>opt; if(opt[0] == 'm'){ int l, r, val; scanf("%d%d%d",&l,&r,&val); Tr.modify(val, l, r); } else { int x; scanf("%d",&x); cout<<Tr.query(x)<<endl; } } }
第二题：一道巨恶心的线段树取模，一般公式知道了就没问题，但lazy乘起来是long long ,一堆人看了一晚上。。。第二天又忘了long long
#include<bits/stdc++.h> using namespace std; #define maxn 100005 #define ll long long int n, m, a[maxn]; int mod = 1000000007; struct SegmentTree{ struct node{ ll sum,lazy[2]; }; node Tree[maxn << 2]; #define ls l, m, v << 1 #define rs m+1, r, v << 1 | 1 void updata(int v){ Tree[v].sum = ( Tree[v << 1].sum % mod + Tree[(v << 1) | 1].sum % mod ) % mod; } void push_down(int l, int r, int v){ if(Tree[v].lazy[0] == 1 && Tree[v].lazy[1] == 0)return; int m = (l + r) >> 1; int A = Tree[v].lazy[0], B = Tree[v].lazy[1]; modify(A, B, l, m, ls); modify(A, B, m+1, r, rs); Tree[v].lazy[0] = 1, Tree[v].lazy[1] = 0; } void build(int l = 1, int r = n, int v = 1){ if(l == r) Tree[v].sum = a[l], Tree[v].lazy[0] = 1, Tree[v].lazy[1] = 0; else { int m = (l + r) >> 1; build(ls); build(rs); updata(v); Tree[v].lazy[0] = 1, Tree[v].lazy[1] = 0; } } void modify(int A, int B, int L, int R, int l = 1, int r = n, int v = 1){ if(l >= L && r <= R){ ll q = Tree[v].sum; Tree[v].sum = (q * (A % mod) + 1LL * (B % mod) * (r - l + 1 )) % mod; ll olda = Tree[v].lazy[0], oldb = Tree[v].lazy[1]; Tree[v].lazy[0] = A * olda % mod , Tree[v].lazy[1] = ( A * oldb + B )% mod; //printf("olda = %d A = %d newa = %d ",olda,A, Tree[v].lazy[0]); } else { push_down(l, r, v); int m = (l + r) >> 1; if(L <= m)modify(A, B, L, R, ls); if(R > m)modify(A, B, L, R, rs); updata(v); } } ll query(int L,int R,int l = 1, int r = n, int v = 1){ if(l >= L && r <= R)return Tree[v].sum ; else { push_down(l, r, v); int m = (l + r) >> 1; ll ans = 0; if(L <= m) ans = (ans + query(L, R, ls) % mod) % mod; if(R > m)ans = (ans % mod + query(L, R, rs)% mod) % mod; return ans; } } }; SegmentTree Tr; int main(){ freopen("linear.in","r",stdin); freopen("linear.out","w",stdout); cin>>n; for(int i = 1; i <= n; i++)scanf("%d", a + i); Tr.build(); cin>>m; for(int i = 1; i <= m; i++){ string opt; cin>>opt; if(opt[0] == 'm'){ int l, r, A, B; scanf("%d%d%d%d",&l,&r,&A,&B); Tr.modify(A, B, l, r); } else { int l, r; scanf("%d%d",&l,&r); cout<<Tr.query(l, r)<<endl; } } }
第三题 裸的splay扳子，很少找到卿学姐有这么接地气的板子，就搬过了，但splay的地方还是写挂了。。。
#include<bits/stdc++.h> using namespace std; #define maxn 100005 #define ll long long int n, m, a[maxn], val[maxn]; struct SplayTree{ int siz[maxn], fa[maxn], son[maxn][2]; int tot, root; void update(int nd){ siz[nd] = siz[son[nd][0]] + siz[son[nd][1]] + 1; } void init(){ siz[0] = 0; tot = 0; root = build(0, 1, n); } int build(int f, int l,int r){ if(l > r)return 0; int nd = ++tot; fa[nd] = f; int m = (l + r) >> 1; son[nd][0] = build(nd, l, m - 1); son[nd][1] = build(nd, m + 1, r); val[nd] = a[m]; update(nd); //printf("nd = %d s[0] = %d, s[1] = %d, l = %d, r = %d val = %d ",nd,son[nd][0],son[nd][1],l,r,val[nd]); return nd; } int find(int pos){ int nd = root; while(1){ int ls = siz[son[nd][0]]; if(pos <= ls)nd = son[nd][0]; else if(pos >= ls + 2){ pos -= ls + 1; nd = son[nd][1]; } else return nd; } } int query(int pos){ return val[find(pos)]; } void rotate(int x,int d){ int y = fa[x]; son[y][d^1] = son[x][d]; if(son[x][d])fa[son[x][d]] = y; fa[x] = fa[y]; if(fa[y]){ if(y == son[fa[y]][d])son[fa[y]][d] = x; else son[fa[y]][d^1] = x; } son[x][d] = y, fa[y] = x; update(y), update(x); } void splay(int x,int target = 0){ while(fa[x] != target){ int y = fa[x]; if(x == son[y][0]){ if(fa[y] != target && y == son[fa[y]][0]) rotate(y, 1); rotate(x, 1); } else { if(fa[y] != target && y == son[fa[y]][1]) rotate(y, 0); rotate(x, 0); } } if(!target)root = x; } void erase(int pos){ int lnd = find(pos - 1), rnd = find(pos + 1); splay(lnd); splay(rnd, lnd); son[rnd][0] = 0; update(rnd), update(lnd); } }; SplayTree Tr; int main(){ freopen("sequence.in","r",stdin); freopen("sequence.out","w",stdout); cin>>n>>m; for(int i = 2; i <= n + 1; i++)scanf("%d", a + i); a[1] = 0, a[n + 2] = 0; n += 2; Tr.init(); for(int i = 1; i <= m; i++){ string opt; int pos; cin>>opt; scanf("%d",&pos); if(opt[0] == 'D')Tr.erase(pos + 1); else printf("%d ",Tr.query(pos + 1)); } }