spoj 2713 线段树

给定一个序列，有两种操作：对一个区间内的数字开方和求区间内所有数字的和。注意到一个即使很大的数经过没几次开方操作以后就会变成1，而1开方还是1。所以可以用线段树来维护，对于那些全部都是1的区间（即区间和等于区间长度）我们不用更新，剩下的就是区间求和了。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;

typedef long long ll;
const int N = 100001;
ll a[N];
int n, m;

struct Node 
{
    int l, r;
    ll sum;
} node[N << 2];

void pushup( int i )
{
    node[i].sum = node[i << 1].sum + node[i << 1 | 1].sum;
}

void build( int i, int l, int r )
{
    node[i].l = l, node[i].r = r;
    if ( l == r )
    {
        node[i].sum = a[l];
        return ;
    }
    int mid = ( l + r ) >> 1;
    build( i << 1, l, mid );
    build( i << 1 | 1, mid + 1, r );
    pushup(i);
}

void update( int i, int l, int r )
{
    int len = node[i].r - node[i].l + 1;
    if ( node[i].sum == len ) return ;
    int mid = ( node[i].l + node[i].r ) >> 1;
    if ( node[i].l == l && node[i].r == r )
    {
        if ( len == 1 )
        {
            node[i].sum = sqrt( node[i].sum * 1.0 );
        }
        else
        {
            update( i << 1, l, mid );
            update( i << 1 | 1, mid + 1, r );
            pushup(i);
        }
        return ;
    }
    if ( r <= mid )
    {
        update( i << 1, l, r );
    }
    else if ( l > mid )
    {
        update( i << 1 | 1, l, r );
    }
    else
    {
        update( i << 1, l, mid );
        update( i << 1 | 1, mid + 1, r );
    }
    pushup(i);
}

ll query( int i, int l, int r )
{
    if ( node[i].l == l && node[i].r == r )
    {
        return node[i].sum;
    }
    int mid = ( node[i].l + node[i].r ) >> 1;
    if ( r <= mid )
    {
        return query( i << 1, l, r );
    }
    else if ( l > mid )
    {
        return query( i << 1 | 1, l, r );
    }
    else
    {
        return query( i << 1, l, mid ) + query( i << 1 | 1, mid + 1, r );
    }
}

int main ()
{
    int _case = 1;
    while ( scanf("%d", &n) != EOF )
    {
        printf("Case #%d:
", _case++);
        for ( int i = 1; i <= n; i++ )
        {
            scanf("%lld", &a[i]);
        }
        build( 1, 1, n );
        scanf("%d", &m);
        while ( m-- )
        {
            int op, x, y;
            scanf("%d%d%d", &op, &x, &y);
            if ( x > y ) swap( x, y );
            if ( op == 0 )
            {
                update( 1, x, y );
            }
            else
            {
                ll ans = query( 1, x, y );
                printf("%lld
", ans);
            }
        }
        putchar('
');
    }
    return 0;
}