BZOJ 3211: 花神游历各国( 线段树 )

线段树...区间开方...明显是要处理到叶节点的

之前在CF做过道区间取模...差不多, 只有开方, 那么每个数开方次数也是有限的(0,1时就会停止), 最大的数10^9开方10+次也就不会动了.那么我们线段树多记个max就可以少掉很多不必要的操作

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#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<iostream> #define rep(i, n) for(int i = 0; i < n; i++)#define M(l, r) (((l) + (r)) >> 1)#define clr(x, c) memset(x, c, sizeof(x)) using namespace std; typedef long long ll; const int maxn = 100009; struct Node { Node *l, *r; int v, mx; ll sum; Node():mx(0) { l = r = NULL; } inline void update() { if(l) { mx = max(l->mx, r->mx); sum = l->sum + r->sum; } else sum = mx = v; }} pool[maxn << 1], *pt = pool, *root; int L, R, n, seq[maxn]; void build(Node* t, int l, int r) { if(r > l) { int m = M(l, r); build(t->l = pt++, l, m); build(t->r = pt++, m + 1, r); } else t->v = seq[l]; t->update();} void modify(Node* t, int l, int r) { if(t->mx <= 1) return; if(l == r) t->v = floor(sqrt(t->v)); else { int m = M(l, r); if(L <= m) modify(t->l, l, m); if(m < R) modify(t->r, m + 1, r); } t->update();} ll query(Node* t, int l, int r) { if(L <= l && r <= R) return t->sum; int m = M(l, r); return (L <= m ? query(t->l, l, m) : 0 ) + (m < R ? query(t->r, m + 1, r) : 0);} int main() {// freopen("test.in", "r", stdin); cin >> n; for(int i = 1; i <= n; i++) scanf("%d", seq + i); build(root = pt++, 1, n); int m, op; cin >> m; while(m--) { scanf("%d%d%d", &op, &L, &R); if(op == 1) printf("%lld\n", query(root, 1, n)); else modify(root, 1, n); } return 0;}--------------------------------------------------------------------------------------------

3211: 花神游历各国

Time Limit: 5 Sec Memory Limit: 128 MB
Submit: 1414 Solved: 546
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Description

Input

Output

每次x=1时，每行一个整数，表示这次旅行的开心度

Sample Input

4

1 100 5 5

5

1 1 2

2 1 2

1 1 2

2 2 3

1 1 4

Sample Output

101

11

11

HINT

对于100%的数据， n ≤ 100000，m≤200000 ,data[i]非负且小于10^9

Source

SPOJ2713 gss4 数据已加强