专题测试二 树形结构 B - The Child and Sequence

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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  .
   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.

   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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