CodeForces 445E DZY Loves Colors

DZY Loves Colors

Time Limit: 2000ms

Memory Limit: 262144KB

This problem will be judged on CodeForces. Original ID: 445E
64-bit integer IO format: %I64d Java class name: (Any)

DZY loves colors, and he enjoys painting.

On a colorful day, DZY gets a colorful ribbon, which consists of n units (they are numbered from 1 to n from left to right). The color of the i-th unit of the ribbon is i at first. It is colorful enough, but we still consider that the colorfulness of each unit is 0 at first.

DZY loves painting, we know. He takes up a paintbrush with color x and uses it to draw a line on the ribbon. In such a case some contiguous units are painted. Imagine that the color of unit i currently is y. When it is painted by this paintbrush, the color of the unit becomes x, and the colorfulness of the unit increases by |x - y|.

DZY wants to perform m operations, each operation can be one of the following:

Paint all the units with numbers between l and r (both inclusive) with color x.
Ask the sum of colorfulness of the units between l and r (both inclusive).

Can you help DZY?

Input

The first line contains two space-separated integers n, m (1 ≤ n, m ≤ 105).

Each of the next m lines begins with a integer type (1 ≤ type ≤ 2), which represents the type of this operation.

If type = 1, there will be 3 more integers l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 108) in this line, describing an operation 1.

If type = 2, there will be 2 more integers l, r (1 ≤ l ≤ r ≤ n) in this line, describing an operation 2.

Output

For each operation 2, print a line containing the answer — sum of colorfulness.

Sample Input

Input

Output

Input

Output

3
2
1

Input

Output

Hint

In the first sample, the color of each unit is initially [1, 2, 3], and the colorfulness is [0, 0, 0].

After the first operation, colors become [4, 4, 3], colorfulness become [3, 2, 0].

After the second operation, colors become [4, 5, 5], colorfulness become [3, 3, 2].

So the answer to the only operation of type 2 is 8.

Source

Codeforces Round #254 (Div. 2)

解题：线段树

 #include <bits/stdc++.h>

 using namespace std;

 typedef long long LL;

 const int maxn = ;

 struct node{

     int lt,rt,color;

     LL sum,len,add;

 }tree[maxn<<];

 void pushup(int v){

     tree[v].sum = tree[v<<].sum + tree[v<<|].sum;

     if(tree[v<<].color == tree[v<<|].color)

         tree[v].color = tree[v<<].color;

     else tree[v].color = ;

 }

 void pushdown(int v){

     if(tree[v].add) {

         tree[v<<].add += tree[v].add;

         tree[v<<|].add += tree[v].add;

         tree[v<<].sum += tree[v].add*tree[v<<].len;

         tree[v<<|].sum += tree[v].add*tree[v<<|].len;

         tree[v].add = ;

     }

     if(tree[v].color){

         tree[v<<].color = tree[v<<|].color = tree[v].color;

         tree[v].color = ;

     }

 }

 void build(int lt,int rt,int v){

     tree[v].lt = lt;

     tree[v].rt = rt;

     tree[v].len = rt - lt + ;

     tree[v].add = ;

     tree[v].sum = ;

     if(lt == rt){

         tree[v].color = lt;

         return;

     }

     int mid = (lt + rt)>>;

     build(lt,mid,v<<);

     build(mid+,rt,v<<|);

     pushup(v);

 }

 void update(int lt,int rt,int color,int v){

     if(lt <= tree[v].lt && rt >= tree[v].rt && tree[v].color){

         tree[v].add += abs(tree[v].color - color);

         tree[v].sum += abs(tree[v].color - color)*tree[v].len;

         tree[v].color = color;

         return;

     }

     pushdown(v);

     if(lt <= tree[v<<].rt) update(lt,rt,color,v<<);

     if(rt >= tree[v<<|].lt) update(lt,rt,color,v<<|);

     pushup(v);

 }

 LL query(int lt,int rt,int v){

     if(lt <= tree[v].lt && rt >= tree[v].rt) return tree[v].sum;

     pushdown(v);

     LL sum = ;

     if(lt <= tree[v<<].rt) sum += query(lt,rt,v<<);

     if(rt >= tree[v<<|].lt) sum += query(lt,rt,v<<|);

     pushup(v);

     return sum;

 }

 int main(){

     int n,m,op,x,y,color;

     scanf("%d%d",&n,&m);

     build(,n,);

     while(m--){

         scanf("%d%d%d",&op,&x,&y);

         if(op == ){

             scanf("%d",&color);

             update(x,y,color,);

         }else printf("%I64d\n",query(x,y,));

     }

     return ;

 }