CodeForces 438D The Child and Sequence

题目链接：http://codeforces.com/problemset/problem/438/D

The Child and Sequence

time limit per test ：4 seconds

memory limit per test ：256 megabytes

input ：standard input

output ：standard output

At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at him. A lot of important things were lost, in particular the favorite sequence of Picks.

Fortunately, Picks remembers how to repair the sequence. Initially he should create an integer array a[1], a[2], ..., a[n]. Then he should perform a sequence of m operations. An operation can be one of the following:

1. Print operation l, r. Picks should write down the value of .
2. Modulo operation l, r, x. Picks should perform assignment a[i] = a[i] mod x for each i (l ≤ i ≤ r).
3. Set operation k, x. Picks should set the value of a[k] to x (in other words perform an assignment a[k] = x).

Can you help Picks to perform the whole sequence of operations?

Input

The first line of input contains two integer: n, m (1 ≤ n, m ≤ 10⁵). The second line contains n integers, separated by space: a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ 10⁹) — initial value of array elements.

Each of the next m lines begins with a number type .

• If type = 1, there will be two integers more in the line: l, r (1 ≤ l ≤ r ≤ n), which correspond the operation 1.
• If type = 2, there will be three integers more in the line: l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 10⁹), which correspond the operation 2.
• If type = 3, there will be two integers more in the line: k, x (1 ≤ k ≤ n; 1 ≤ x ≤ 10⁹), which correspond the operation 3.

Output

For each operation 1, please print a line containing the answer. Notice that the answer may exceed the 32-bit integer.

Examples

Input

5 5
1 2 3 4 5
2 3 5 4
3 3 5
1 2 5
2 1 3 3
1 1 3

Output

8
5

Input

10 10
6 9 6 7 6 1 10 10 9 5
1 3 9
2 7 10 9
2 5 10 8
1 4 7
3 3 7
2 7 9 9
1 2 4
1 6 6
1 5 9
3 1 10

Output

49
15
23
1
9

Note

Consider the first testcase:

• At first, a = {1, 2, 3, 4, 5}.
• After operation 1, a = {1, 2, 3, 0, 1}.
• After operation 2, a = {1, 2, 5, 0, 1}.
• At operation 3, 2 + 5 + 0 + 1 = 8.
• After operation 4, a = {1, 2, 2, 0, 1}.
• At operation 5, 1 + 2 + 2 = 5.

思路：稍微分析一下题目，就能够看出这道题是线段树的单点更新，区间更新，区间求和问题。但是有一个问题存在，这里区间更新是取模，直接对和进行取模肯定是错误的，所以得换种想法做。考虑题目给的时间很充足，而且测试数据是单个样例测试，所以可以尝试区间更新时更新到叶子节点。但这种做法好像还是不太保险，仔细斟酌一下，其实对于取模操作，只有当该数大于模数，才需要更新，否则不需要，这样就类似于DFS的剪枝了，所以将每个区间的最大值求出来，更新时与模数比较，模数大就直接返回，不必向下更新了，以节约时间开销，这样就更有胜算了，事实证明是的，后来算了一下复杂度，发现这样做的话复杂度并不是很高。详见代码。

附上AC代码：

#include <bits/stdc++.h>
#define lrt rt<<1
#define rrt rt<<1|1
#define lson l, m, lrt
#define rson m+1, r, rrt
using namespace std;
typedef long long ll;
const int maxn = 100005;
ll sumv[maxn<<2], maxv[maxn<<2];
int n, q;

void push_up(int rt){
	sumv[rt] = sumv[lrt]+sumv[rrt];
	maxv[rt] = max(maxv[lrt], maxv[rrt]);
}

void build(int l, int r, int rt){
	if (l == r){
		int num;
		cin >> num;
		sumv[rt] = num;
		maxv[rt] = num;
		return ;
	}
	int m = (l+r)>>1;
	build(lson);
	build(rson);
	push_up(rt);
}

void update1(int p, int val, int l, int r, int rt){
	if (l == r){
		sumv[rt] = val;
		maxv[rt] = val;
		return ;
	}
	int m = (l+r)>>1;
	if (p <= m)
		update1(p, val, lson);
	else
		update1(p, val, rson);
	push_up(rt);
}

void update2(int cl, int cr, int val, int l, int r, int rt){
	if (maxv[rt] < val)
		return ;
	if (l == r){
		sumv[rt] %= val;
		maxv[rt] %= val;
		return ;
	}
	int m = (l+r)>>1;
	if (cl <= m)
		update2(cl, cr, val, lson);
	if (cr > m)
		update2(cl, cr, val, rson);
	push_up(rt);
}

ll query(int ql, int qr, int l, int r, int rt){
	if (ql<=l && r<=qr)
		return sumv[rt];
	int m = (l+r)>>1;
	ll sumr = 0;
	if (ql <= m)
		sumr += query(ql, qr, lson);
	if (qr > m)
		sumr += query(ql, qr, rson);
	return sumr;
}

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin >> n >> q;
	build(1, n, 1);
	int type, l, r, x;
	while (q--){
		cin >> type >> l >> r;
		if (1 == type)
			cout << query(l, r, 1, n, 1) << endl;
		else if (2 == type){
			cin >> x;
			update2(l, r, x, 1, n, 1);
		}
		else
			update1(l, r, 1, n, 1);
	}
	return 0;
}