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;

}

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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