【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() {
        
        read(n); 
        for (i = 1; i <= n; i++) read(a[i]);
        T.build(1,1,n);
        read(m);
        while (m--)
        {
                read(opt);
                if (opt == 1)
                {
                        read(l); read(r);
                        writeln(T.query(1,l,r));
                } else
                {
                        read(l); read(r);
                        T.modify(1,l,r);
                }
        }
        return 0;
    
}

 

posted @ 2018-06-02 09:34  evenbao  阅读(95)  评论(0编辑  收藏  举报