hdu 4027 Can you answer these queries? The 36th ACM/ICPC Asia Regional Shanghai Site

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=4027

The 36th ACM/ICPC Asia Regional Shanghai Site —— Online Contest

The 36th ACM/ICPC Asia Regional Shanghai Site —— Online Contest

比赛的时候知道必须成段更新但是没有想到怎么更新，于是就单点更新了，果断TLE，其实只要知道__int64 范围内的数最多取8次跟下就能到1或0，这样就有思路了，就是预先把每个数的i次根下都求出来，在建树的过程中直接把和更新上去，这样每次更新只需要更新取根次数就可以了。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>

#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

using namespace std;

const int maxx=100002;

struct Tree
{
    __int64 sum[9];
    int cnt;
    int col;
};

__int64 da[maxx],mx;
int lim;

Tree tree[maxx<<2];

void pushUp(int rt)
{
    for (int i=0;i<=lim;i++)
    {
        tree[rt].sum[i]=tree[rt<<1].sum[i]+tree[rt<<1|1].sum[i];
    }
}

void pushDown(int rt)
{
    if (tree[rt].col!=0)
    {
        tree[rt<<1].cnt+=tree[rt].col;
        tree[rt<<1].col+=tree[rt].col;
        tree[rt<<1|1].cnt+=tree[rt].col;
        tree[rt<<1|1].col+=tree[rt].col;
        tree[rt].col=0;
    }
}

void update(int L,int R,int l,int r,int rt)
{
    if (L<=l && r<=R)
    {
        tree[rt].col+=1;
        tree[rt].cnt+=1;
        return;
    }
    pushDown(rt);
    int m=(l+r)>>1;
    if (L<=m)update(L,R,lson);
    if (R>m)update(L,R,rson);

}

void build(int l,int r,int rt)
{
    tree[rt].cnt=tree[rt].col=0;
    if (l==r)
    {
        tree[rt].sum[0]=da[l];
        for (int i=1;i<=lim;i++)
        {
            tree[rt].sum[i]=(__int64)sqrt((double)tree[rt].sum[i-1]);
        }
        return;
    }
    int m=(l+r)>>1;
    build(lson);
    build(rson);
    pushUp(rt);
}

__int64 query(int L,int R,int l,int r,int rt)
{
    if (L<=l && r<=R)
    {
        if (tree[rt].cnt>lim)tree[rt].cnt=lim;
        if (tree[rt].cnt==lim || l==r)
        {
            return tree[rt].sum[tree[rt].cnt];
        }
        pushDown(rt);
        int m=(l+r)>>1;
        return query(L,R,lson)+query(L,R,rson);
    }
    pushDown(rt);
    int m=(l+r)>>1;
    __int64 ret=0;
    if (L<=m) ret+=query(L,R,lson);
    if (R>m) ret+=query(L,R,rson);
    return ret;
}

int main()
{
    int n,o,cas=1;
    while (scanf("%d",&n)!=EOF)
    {
        mx=0;
        int i;
        printf("Case #%d:\n",cas++);
        for (i=1;i<=n;i++)
        {
            scanf("%I64d",&da[i]);
            if (mx<da[i])mx=da[i];
        }
        for (lim=1;mx>1;mx=(__int64)sqrt((double)mx),lim++);
        build(1,n,1);
        scanf("%d",&o);
        for (i=1;i<=o;i++)
        {
            int x,y,mark;
            scanf("%d%d%d",&mark,&x,&y);
            if (x>y)swap(x,y);
            if (mark==0)
            {
                update(x,y,1,n,1);
            }
            else
            {
                printf("%I64d\n",query(x,y,1,n,1));
            }
        }
        printf("\n");
    }
    return 0;
}

The 36th ACM/ICPC Asia Regional Shanghai Site —— Online Contest

The 36th ACM/ICPC Asia Regional Shanghai Site —— Online Contest

The 36th ACM/ICPC Asia Regional Shanghai Site —— Online Contest

The 36th ACM/ICPC Asia Regional Shanghai Site —— Online Contest