【BZOJ 3211&3038】 花神游历各国 & 上帝造题的七分钟2

【题目链接】

【BZOJ 3211】 点击打开链接

【BZOJ 3038】 点击打开链接

【算法】

线段树

开根操作直接开到叶子节点，注意当区间中所有数都是0或1时，不需要开根

【代码】

#include<bits/stdc++.h>
using namespace std;
#define MAXN 100010

int i,n,m,opt,l,r;
long long a[MAXN];

struct SegmentTree
{
struct Node
{
int l,r;
long long sum;
bool flag;
} Tree[MAXN<<2];
inline void update(int index)
{
Tree[index].sum = Tree[index<<1].sum + Tree[index<<1|1].sum;
Tree[index].flag = Tree[index<<1].flag & Tree[index<<1|1].flag;
}
inline void build(int index,int l,int r)
{
int mid;
Tree[index].l = l; Tree[index].r = r;
if (l == r)
{
Tree[index].sum = a[l];
if (a[l] == 0 || a[l] == 1) Tree[index].flag = true;
else Tree[index].flag = false;
return;
}
mid = (l + r) >> 1;
build(index<<1,l,mid);
build(index<<1|1,mid+1,r);
update(index);
}
inline void modify(int index,int l,int r)
{
int mid;
if (Tree[index].flag) return;
if (Tree[index].l == Tree[index].r)
{
Tree[index].sum = (long long)sqrt(Tree[index].sum);
if (Tree[index].sum == 0 || Tree[index].sum == 1) Tree[index].flag = true;
return;
}
mid = (Tree[index].l + Tree[index].r) >> 1;
if (mid >= r) modify(index<<1,l,r);
else if (mid + 1 <= l) modify(index<<1|1,l,r);
else
{
modify(index<<1,l,mid);
modify(index<<1|1,mid+1,r);
}
update(index);
}
inline long long query(int index,int l,int r)
{
int mid;
if (Tree[index].l == l && Tree[index].r == r) return Tree[index].sum;
mid = (Tree[index].l + Tree[index].r) >> 1;
if (mid >= r) return query(index<<1,l,r);
else if (mid + 1 <= l) return query(index<<1|1,l,r);
else return query(index<<1,l,mid) + query(index<<1|1,mid+1,r);
}
} T;
template <typename T> inline void read(T &x)
{
int f = 1; x = 0;
char c = getchar();
for (; !isdigit(c); c = getchar()) { if (c == '-') f = -f; }
for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
x *= f;
}
template <typename T> inline void write(T x)
{
if (x < 0)
{
putchar('-');
x = -x;
}
if (x > 9) write(x/10);
putchar(x%10+'0');
}
template <typename T> inline void writeln(T x)
{
write(x);
puts("");
}

int main() {

for (i = 1; i <= n; i++) read(a[i]);
T.build(1,1,n);
while (m--)
{
if (opt == 1)
{
writeln(T.query(1,l,r));
} else
{
}