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 betweenlandr(both inclusive) with colorx.
- Ask the sum of colorfulness of the units betweenlandr(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
//
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;
}
