Codeforces Round #250 (Div. 1) D. The Child and Sequence

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.

Sample test(s)

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.

• /*
  多校被虐>_<。一些题题解都看不懂，只有默默来写CF了。。。
  题意很好理解，
  解法：对于维护区间最大值mx和区间sum，当 x > mx时不再更新下去，
  否则一直更新到根节点
  总结:遇到一个居间的数按照某种方式改变时，我们更新的时候，要抓住什么
  什么不更新下去是不影响最后结果的，就像本题，如果孩子的最大值mx < x
  更新是没有意思的。
  */
  
  #include<cstdio>
  #include<iostream>
  #include<algorithm>
  #include<cstring>
  #include<cmath>
  #include<queue>
  #include<ctime>
  #include<map>
  #include<set>
  #include<stack>
  #include<list>
  #define maxn 100010
  #define LL long long
  using namespace std ;
  
  struct node
  {
      LL sum ;
      int mx,mod ;
  }tree[maxn*3];
  int ql ,qr;
  LL v ;
  void pushup( int L ,int R,int o)
  {
      tree[o].mx = max(tree[o<<1].mx,tree[o<<1|1].mx) ;
      tree[o].sum = tree[o<<1].sum+tree[o<<1|1].sum ;
      tree[o].mod = 0 ;
  }
  void insert( int L ,int R ,int o )
  {
      if(L==R)
      {
          tree[o].sum = v;
          tree[o].mx = v ;
          tree[o].mod = 0 ;
          return ;
      }
      int mid = (L+R)>>1 ;
      if(ql <=mid) insert(L,mid,o<<1) ;
      else insert(mid+1,R,o<<1|1) ;
      pushup(L,R,o) ;
  }
  void update( int L,int R,int o)
  {
      if(L==R)
      {
          tree[o].mx %= v ;
          tree[o].sum = tree[o].mx;
          return ;
      }
      int mid = (L+R)>>1 ;
      if(ql <= mid && tree[o<<1].mx >= v) update(L,mid,o<<1) ;
      if(qr > mid && tree[o<<1|1].mx >= v ) update(mid+1,R,o<<1|1) ;
      pushup(L,R,o) ;
  }
  void find( int L ,int R ,int o )
  {
      if(ql <= L && qr >= R)
      {
          v += tree[o].sum ;
          return ;
      }
      int mid = (L+R)>>1 ;
      if(ql <= mid) find(L,mid,o<<1) ;
      if(qr > mid) find(mid+1,R,o<<1|1) ;
      pushup(L,R,o) ;
  }
  int main()
  {
      int i , n , m ,j , k;
      while( scanf("%d%d",&n,&m) != EOF)
      {
          for( ql = 1 ; ql <= n ;ql++)
          {
              scanf("%I64d",&v) ;
              insert(1,n,1) ;
          }
          while(m--)
          {
              scanf("%d",&k) ;
              if(k==1)
              {
                  scanf("%d%d",&ql,&qr) ;
                  v = 0 ;
                  find(1,n,1) ;
                  printf("%I64d\n",v) ;
              }
              else if(k==2)
              {
                  scanf("%d%d%I64d",&ql,&qr,&v) ;
                  update(1,n,1) ;
              }
              else{
                  scanf("%d%I64d",&ql,&v) ;
                  insert(1,n,1) ;
              }
          }
      }
      return 0 ;
  }