传送门
这道题看一眼就知道是线段树的题目,但是一看,题目要求区间开方,我就又傻了。想了一会,发现就算是109在开方5次之后就变成1了,所以就算是我们单个开方,时间复杂度也就是O(NlogN)。但是为了避免重复开方,我们要给线段树的每一个节点做一个标记,表示下面的区间是否全部为10。那样我们就可以避免重复开方了。

代码:

#include <cstdio>
#include <cmath>
#define LL long long
inline void swap(int &a, int &b) {int c = a; a = b; b = c;}
#define MAXN 100005
inline void GET(LL &n)
{
    n = 0; char c;
    do c = getchar(); while(c > '9' || c < '0');
    while(c <= '9' && c >= '0') {n = n * 10 + c - '0'; c = getchar();}
}
inline void GET(int &n)
{
    n = 0; char c;
    do c = getchar(); while(c > '9' || c < '0');
    while(c <= '9' && c >= '0') {n = n * 10 + c - '0'; c = getchar();}
}
struct node
{
    LL sum;
    bool lazy;
    node(){lazy = 0;}
}t[MAXN<<1];
inline int idx(int l, int r) {return (l + r) | (l != r);}
void pushup(int l, int r)
{
    int mid = (l + r) >> 1;
    t[idx(l, r)].sum = t[idx(l, mid)].sum + t[idx(mid+1, r)].sum;
    t[idx(l, r)].lazy = t[idx(l, mid)].lazy && t[idx(mid+1, r)].lazy;
}
void build(int l, int r)
{
    int i = idx(l, r);
    if(l == r)
    {
        GET(t[i].sum);
        return;
    }
    int mid = (l + r) >> 1;
    build(l, mid); build(mid+1, r);
    t[idx(l, r)].sum = t[idx(l, mid)].sum + t[idx(mid+1, r)].sum;
}
int op, L, R, n, m;
void Ins(int l, int r)
{
    int i = idx(l, r);
    if(l > R || r < L) return;
    if(t[i].lazy) return;
    if(l == r)
    {
        t[i].sum = sqrt(t[i].sum);
        if(t[i].sum == 1 || t[i].sum == 0) t[i].lazy = 1;
        return;
    }
    int mid = (l + r) >> 1;
    Ins(l, mid);
    Ins(mid+1, r);
    pushup(l, r);
}
LL Sum(int l, int r)
{
    if(l > R || r < L) return 0;
    if(l >= L && r <= R) return t[idx(l, r)].sum;
    int mid = (l+r)>>1;
    return Sum(l, mid) + Sum(mid+1, r);
}
int main()
{
    GET(n);
    build(1, n);
    GET(m);
    while(m --)
    {
        GET(op);GET(L);GET(R);
        if(op == 2)
            Ins(1, n);
        else if(op == 1)
            printf("%lld\n", Sum(1, n));
    }
    return 0;
}
posted on 2015-08-23 13:47  geng4512  阅读(102)  评论(0编辑  收藏  举报