区间专题

int mx
[20];                      //最多能保存524288的长度
int RMQ(int l, int r)   {
int k = 0;  while (1<<(k+1) <= r - l + 1)   k++;        //令k为满足1<<k <= r-l+1的最大整数
return max (mx[l][k], mx[r-(1<<k)+1][k]);               //区间最左边1<<k长度的最大值和最右边1<<k长度的最大值,可能有重叠
}

int main(void)    {
for (int i=1; i<=n; ++i)    {
scanf ("%d", &mx[i][0]);
}
for (int j=1; (1<<j)<=n; ++j)    {
for (int i=1; i+(1<<j)-1<=n; ++i)   {
mx[i][j] = max (mx[i][j-1], mx[i+(1<<(j-1))][j-1]);     //从i开始，长度1<<j的区间的最大值
}
}
printf ("%d\n", RMQ (ql, qr));
｝

struct ST    {
int mx[N<<2];
void push_up(int rt)    {
mx[rt] = max (mx[rt<<1], mx[rt<<1|1]);
}
void build(int l, int r, int rt)    {
if (l == r)    {
scanf ("%d", &mx[rt]);    return ;
}
int mid = (l + r) >> 1;
build (lson);　　build (rson);
push_up (rt);
}
int query(int ql, int qr, int l, int r, int rt)    {
if (ql <= l && r <= qr)    {
return mx[rt];
}
int mid = (l + r) >> 1, ret = 0;
if (ql <= mid)    ret = max (ret, query (ql, qr, lson));
if (qr > mid)    ret = max (ret, query (ql, qr, rson));
return ret;
}
}st;