Codeforces Round #254 (Div. 1) C. DZY Loves Colors

C. DZY Loves Colors

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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.

Examples

Input

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

Output

8

Input

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

Output

3
2
1

Input

10 6
1 1 5 3
1 2 7 9
1 10 10 11
1 3 8 12
1 1 10 3
2 1 10

Output

129

Note
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.

【题意】
给定一个长度为n的序列,初始时ai=i,vali=0(1≤i≤n).有两种操作:

将区间[L,R]的值改为x,并且当一个数从y改成x时它的权值vali会增加|x−y|.
询问区间[L,R]的权值和.
n≤105,1≤x≤106.

【解题方法】不明觉厉

【代码君】

//
//Created by just_sort 2016/10/3 14:10
//Copyright (c) 2016 just_sort.All Rights Reserved
//

#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 1e5+7;
typedef long long LL;
struct node{
int l,r,same;
LL tag,sum;
int len(){
return r-l+1;
}
}T[maxn*4];
void pushup(int rt)
{
T[rt].sum = T[rt*2].sum + T[rt*2+1].sum;
if(T[rt*2].same != T[rt*2+1].same) T[rt].same = -1;
else T[rt].same = T[rt*2].same;
}
void update(LL val,int col,int rt)
{
T[rt].tag += val;
T[rt].sum += 1LL*val*T[rt].len();
T[rt].same = col;
}
void pushdown(int rt)
{
if(T[rt].tag)
{
update(T[rt].tag,T[rt].same,rt*2);
update(T[rt].tag,T[rt].same,rt*2+1);
T[rt].tag = 0;
}
}
void Build(int l,int r,int rt)
{
T[rt].l = l,T[rt].r = r,T[rt].same = -1;
T[rt].sum = T[rt].tag = 0;
if(l == r){
T[rt].same = l;
}
else{
int m = (l + r)/2;
Build(l,m,rt*2);
Build(m+1,r,rt*2+1);
}
}
void modify(int L,int R,int val,int rt)
{
if(L == T[rt].l && T[rt].r == R && ~T[rt].same)
{
update(abs(val-T[rt].same),val,rt);
}
else{
pushdown(rt);
int mid = (T[rt].l + T[rt].r)/2;
if(R <= mid) modify(L,R,val,rt*2);
else if(L > mid) modify(L,R,val,rt*2+1);
else{
modify(L,mid,val,rt*2);
modify(mid+1,R,val,rt*2+1);
}
pushup(rt);
}
}
LL queryans(int L,int R,int rt)
{
if(T[rt].l == L && T[rt].r == R)
{
return T[rt].sum;
}
else{
pushdown(rt);
int mid = (T[rt].l + T[rt].r)/2;
if(R <= mid) return queryans(L,R,rt*2);
else if(L > mid) return queryans(L,R,rt*2+1);
else{
LL ans = 0;
ans += queryans(L,mid,rt*2);
ans += queryans(mid+1,R,rt*2+1);
return ans;
}
}
}
int main()
{
int n,m;
scanf("%d%d",&n,&m);
Build(1,n,1);
int op,l,r,x;
for(int i = 1; i <= m; i++)
{
scanf("%d",&op);
if(op == 1)
{
scanf("%d%d%d",&l,&r,&x);
modify(l,r,x,1);
}
else{
scanf("%d%d",&l,&r);
printf("%lld\n",queryans(l,r,1));
}
}
return 0;
}