HDOJ 4027 Can you answer these queries?

此题陷阱多多：数据类型（相应的输入输出）、区间端点可能反序、如果用C提交注意强制转换的格式（int(x)是不对的）；

思路：每次如果开根号的区间内含有未达到0或1的数就一直向下，找到这些数并直接更新一次，这样最多跟新4*N次，加上线段树的复杂度，为4*N*lgN，是最坏情况的复杂度，可能由于开根号耗时较多，总体比较耗时（700ms+），查询的复杂度为lgN。

# include <stdio.h>

# include <math.h>



# define N 100005



# define ls ((r)<<1)

# define rs ((r)<<1|1)

# define mid (((x)+(y))>>1)



typedef long long int LL;



LL n, m;

char bott[N<<2];

LL sum[N<<2];



LL root(LL x)

{

        return (LL)floor(sqrt(x));

}



void update(LL r)

{

    sum[r] = sum[ls] + sum[rs];

    bott[r] = bott[ls] & bott[rs];

}



void build(LL r, LL x, LL y)

{

    bott[r] = 0;

    if (x == y)

    {

        scanf("%I64d", &sum[r]);

        return ;

    }

    build(ls, x, mid);

    build(rs, mid+1, y);

    update(r);

}



void srt(LL r, LL x, LL y, LL s, LL t)

{

    if (s<=x && y<=t)

    {

        if (bott[r]) return ;

        else if (x == y)

        {

            sum[r] = root(sum[r]);

            if (sum[r] < 2) bott[r] = 1;

            return ;

        }

    }

    if (s <= mid) srt(ls, x, mid, s, t);

    if (mid+1 <= t) srt(rs, mid+1, y, s, t);

    update(r);

}



void query(LL r, LL x, LL y, LL s, LL t, LL *ans)

{

    if (s<=x && y<=t)

    {

        *ans += sum[r];

        return ;

    }

    if (s<=mid) query(ls, x, mid, s, t, ans);

    if (mid+1<=t) query(rs, mid+1, y, s, t, ans);

}



void solve(void)

{

    LL i, op, s, t, tmp; LL ans;

    scanf("%I64d", &m);

    for (i = 1; i <= m; ++i)

    {

        scanf("%I64d%I64d%I64d", &op, &s, &t);

        //printf("%I64d %I64d %I64d\n", op, s, t);

        if (s > t) tmp = s, s = t, t = tmp;

        if (op == 0)    srt(1, 1, n, s, t);

        else

        {

            ans = 0;

            query(1, 1, n, s, t, &ans);

            printf("%I64d\n", ans);

        }

    }

}



int main()

{

    LL icase = 0;

    while (~scanf("%I64d", &n))

    {

        printf("Case #%I64d:\n", ++icase);

        build(1, 1, n);

        solve();

        putchar('\n');

    }



    return 0;

}