链接：https://ac.nowcoder.com/acm/contest/26896/1001
来源：牛客网

题目描述

For the daily milking, Farmer John's N cows (1 ≤ N ≤ 100,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate Frisbee with some of the cows. To keep things simple, he will take a contiguous range of cows from the milking lineup to play the game. However, for all the cows to have fun they should not differ too much in height.
Farmer John has made a list of Q (1 ≤ Q ≤ 30) potential groups of cows and their heights (1 ≤ height ≤ 1,000,000). For each group, he wants your help to determine the difference in height between the shortest and the tallest cow in the group.

输入描述:

Line 1: Two space-separated integers, N and Q.
Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i
Lines N+2..N+Q+1: Two integers A and B (1 ≤ A ≤ B ≤ N), representing the range of cows from A to B inclusive.

输出描述:

Lines 1..Q: Each line contains a single integer that is a response to a reply and indicates the difference in height between the tallest and shortest cow in the range.

示例1

输入

复制 6 3 1 7 3 4 2 5 1 5 4 6 2 2

6 3
1
7
3
4
2
5
1 5
4 6
2 2

输出

复制 6 3 0

6
3
0

分析

易错点：这种两个小于号的写法，是所求区间不变，查询区间挨个变小，比较的时候也要用查询区间和所求区间比较，不然很容易出bug。。

//-------------------------代码----------------------------

//#define int ll
const int N = 1e5+10;
int n,m;
int a[N];

int tr1[4*N];
int tr2[4*N];

void init(int p,int l,int r) {
    if(l == r)  {
        tr1[p] = a[l];
        tr2[p] = a[l];
        rt;
    }
    int mid = l + r >>1;
    init(2 * p,l,mid);
    init(2 * p + 1,mid + 1,r);
    tr1[p] = max(tr1[p * 2],tr1[p * 2 + 1]); 
    tr2[p] = min(tr2[p * 2],tr2[p * 2 + 1]);
    
}

int calc(int p,int x,int y,int l,int r,bool q) {
    if(x <= l && r <= y ) {
        if(q == 1)return tr1[p];
        else return tr2[p];
    } 
    int mid = l + r >> 1;
    int mx = 0;
    int mn = 1e9;
    if(x <= mid) {
        if(q)mx = max(mx , calc(p << 1, x, y, l, mid,1));
        if(!q)mn = min(mn , calc(p << 1, x, y, l, mid,0));
    }
    if(y > mid ) {
        if(q)mx = max(mx , calc(p << 1 | 1, x, y, mid + 1, r,1));
        if(!q)mn = min(mn , calc(p << 1 | 1, x, y, mid + 1, r,0));
    }
    if(q == 1) return mx;
    else return mn;
}


void solve()
{
//    cin>>n>>m;
    cin>>n>>m;
    fo(i,1,n) {
        cin>>a[i];
    }
    init(1,1,n);
    fo(i,1,m) {
        int l,r;
        cin>>l>>r;
        cout<<calc(1,l,r,1,n,1) - calc(1,l,r,1,n,0)<<endl;
    }
}
void main_init() {}
signed main(){
    AC();clapping();TLE;
    cout<<fixed<<setprecision(12);
    main_init();
//  while(cin>>n,n)
//  while(cin>>n>>m,n,m)
//    int t;cin>>t;while(t -- )
    solve();
//    {solve(); }
    return 0;
}

/*样例区


*/

//------------------------------------------------------------