专题测试二 树形结构 B - The Child and Sequence

  1. 题目

    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 专题测试二 树形结构 B - The Child and Sequence .
    2. Modulo operation l, r, x. Picks should perform assignment a[i] = a[imod 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.

    Sample 1
    Input Output
    5 5
    1 2 3 4 5
    2 3 5 4
    3 3 5
    1 2 5
    2 1 3 3
    1 1 3
    
    8
    5
    
    Sample 2
    Inputcopy Outputcopy
    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
    
    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.
  2. 思路
    线段树,操作有区间求和,区间取模,单点修改
    每次全部取模的话必定超时,所以要剪枝,因为每次取模之后数组里的数要么不变要么变小,除非修改之后才有可能变大,可以存储每个节点代表的数组区间中元素的最大值,如果最大的数都比模数小,很明显这个区间就没必要继续模了,直接跳过。数据强度似乎不大,这么剪就能直接过了。
  3. 代码
    #include<cstdio>
    #include<iostream>
    #include<string>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    
    #define ll long long
    
    ll d[1000000],a[1000005],maxx[1000005];//区间值,原始值
    ll n,q,mod;
    
    void build(ll s, ll t, ll p) {
    
    	if (s == t) {
    	    maxx[p] = d[p] = a[s];
    	    return;
    	}
    
    	ll m = s + ((t - s) >> 1);
    	build(s, m, p * 2), build(m + 1, t, p * 2 + 1);
    	d[p] = d[p * 2] + d[(p * 2) + 1];
    	maxx[p]=max(maxx[p * 2],maxx[(p * 2) + 1]);
    }
    
    void update(ll l, ll c, ll s, ll t, ll p) {
    
    	if (l == s && t == l) {
    		d[p] = c;
    		maxx[p] = c;
    	    return;
    	}
    
    	ll m = s + ((t - s) >> 1);
    	if (l <= m) update(l, c, s, m, p * 2);
    	else update(l, c, m + 1, t, p * 2 + 1);
    	d[p] = d[p * 2] + d[p * 2 + 1];
    	maxx[p]=max(maxx[p * 2],maxx[(p * 2) + 1]);
    }
    
    void updatemo(int l,int r,int mod,int s,int t,int p)
    {
        if(maxx[p] < mod) return;
        if(s == t)
        {
            d[p] %= mod;
            maxx[p] = d[p];
            return;
        }
        int m = s + ((t - s) >> 1);
        if (l <= m) updatemo(l, r, mod, s, m, p * 2);
        if (r > m) updatemo(l, r, mod, m + 1, t, p * 2 + 1);
        d[p] = d[p * 2] + d[(p * 2) + 1];
    	maxx[p]=max(maxx[p * 2],maxx[(p * 2) + 1]);
    }
    
    ll getsum(ll l, ll r, ll s, ll t, ll p) {
    
    	if (l <= s && t <= r) return d[p];
    
    	ll m = s + ((t - s) >> 1);
    	ll sum = 0;
    	if (l <= m) sum += getsum(l, r, s, m, p * 2);
    	if (r > m) sum += getsum(l, r, m + 1, t, p * 2 + 1);
    	return sum;
    }
    
    int main()
    {
    	scanf("%lld%lld",&n,&q);
    	for(int i=1;i<=n;i++){
    		scanf("%lld",&a[i]);
    	}
    	build(1,n,1);
    	ll op,x,y,mod;
    	while (q--) {
    		cin >> op;
    	if (op == 1)
    		cin >> x >> y,cout << getsum(x, y, 1, n, 1) <<endl;
    	else if(op == 2)
    		cin >> x >> y >> mod, updatemo(x, y, mod, 1, n, 1);
    	else
    		cin >> x >> y,update(x, y, 1, n, 1);
    	}
    }
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