BZOJ 3038: 上帝造题的七分钟2 / BZOJ 3211: 花神游历各国 (线段树区间开平方)

题意

给出一些数，有两种操作。(1)将区间内每一个数开方(2)查询每一段区间的和

分析

普通的线段树保留修改+开方优化。可以知道当一个数为0或1时，无论开方几次，答案仍然相同。所以设置flag=1变表示这一段区间全是0/1,那么修改的时候直接暴力遍历线段树结点。因为一个数被开方下取整到1,只会被开方几次,所以说这样修改是O(n)O(n)O(n)的.总时间还是O(nlogn)O(nlogn)O(nlogn)

CODE1

上帝造题的七分钟2

#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
char cb[1<<15],*cs=cb,*ct=cb;
#define getc() (cs==ct&&(ct=(cs=cb)+fread(cb,1,1<<15,stdin),cs==ct)?0:*cs++)
template<class T>inline void read(T &res) {
    char ch; int flg = 1; for(;!isdigit(ch=getc());)if(ch=='-')flg=-flg;
    for(res=ch-'0';isdigit(ch=getc());res=res*10+ch-'0'); res*=flg;
}
const int MAXN = 100005;
int n, m;
struct seg {
	LL sum; bool flg;
}t[MAXN<<2];
inline void upd(int i) {
	t[i].sum = t[i<<1].sum + t[i<<1|1].sum;
	t[i].flg = t[i<<1].flg & t[i<<1|1].flg;
}
void build(int i, int l, int r) {
	t[i].flg = 0;
	if(l == r) {
		read(t[i].sum);
		if(t[i].sum <= 1) t[i].flg = 1;
		return;
	}
	int mid = (l + r) >> 1;
	build(i<<1, l, mid);
	build(i<<1|1, mid+1, r);
	upd(i);
}
void modify(int i, int l, int r, int x, int y) {
	if(t[i].flg) return;
	if(l == r) {
		t[i].sum = sqrt(t[i].sum);
		if(t[i].sum <= 1) t[i].flg = 1;
		return;
	}
	int mid = (l + r) >> 1;
	if(x <= mid) modify(i<<1, l, mid, x, y);
	if(y > mid) modify(i<<1|1, mid+1, r, x, y);
	upd(i);
}
LL query(int i, int l, int r, int x, int y) {
	if(x <= l && r <= y) return t[i].sum;
	int mid = (l + r) >> 1;
	if(y <= mid) return query(i<<1, l, mid, x, y);
	else if(x > mid) return query(i<<1|1, mid+1, r, x, y);
	else return query(i<<1, l, mid, x, y) + query(i<<1|1, mid+1, r, x, y);
}
int main () {
	read(n); build(1, 1, n); read(m);
	int x, y, op;
	while(m--) {
		read(op), read(x), read(y);
		if(x > y) swap(x, y); //坑!
		if(!op) modify(1, 1, n, x, y);
		else printf("%lld\n", query(1, 1, n, x, y));
	}
}

CODE 2

花神游历各国

#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
char cb[1<<15],*cs=cb,*ct=cb;
#define getc() (cs==ct&&(ct=(cs=cb)+fread(cb,1,1<<15,stdin),cs==ct)?0:*cs++)
template<class T>inline void read(T &res) {
    char ch; int flg = 1; for(;!isdigit(ch=getc());)if(ch=='-')flg=-flg;
    for(res=ch-'0';isdigit(ch=getc());res=res*10+ch-'0'); res*=flg;
}
const int MAXN = 100005;
int n, m;
struct seg {
	LL sum; bool flg;
}t[MAXN<<2];
inline void upd(int i) {
	t[i].sum = t[i<<1].sum + t[i<<1|1].sum;
	t[i].flg = t[i<<1].flg & t[i<<1|1].flg;
}
void build(int i, int l, int r) {
	t[i].flg = 0;
	if(l == r) {
		read(t[i].sum);
		if(t[i].sum <= 1) t[i].flg = 1;
		return;
	}
	int mid = (l + r) >> 1;
	build(i<<1, l, mid);
	build(i<<1|1, mid+1, r);
	upd(i);
}
void modify(int i, int l, int r, int x, int y) {
	if(t[i].flg) return;
	if(l == r) {
		t[i].sum = sqrt(t[i].sum);
		if(t[i].sum <= 1) t[i].flg = 1;
		return;
	}
	int mid = (l + r) >> 1;
	if(x <= mid) modify(i<<1, l, mid, x, y);
	if(y > mid) modify(i<<1|1, mid+1, r, x, y);
	upd(i);
}
LL query(int i, int l, int r, int x, int y) {
	if(x <= l && r <= y) return t[i].sum;
	int mid = (l + r) >> 1;
	if(y <= mid) return query(i<<1, l, mid, x, y);
	else if(x > mid) return query(i<<1|1, mid+1, r, x, y);
	else return query(i<<1, l, mid, x, y) + query(i<<1|1, mid+1, r, x, y);
}
int main () {
	read(n); build(1, 1, n); read(m);
	int x, y, op;
	while(m--) {
		read(op), read(x), read(y);
		if(op == 2) modify(1, 1, n, x, y);
		else printf("%lld\n", query(1, 1, n, x, y));
	}
}