Transformation

Time Limit: 15000/8000 MS (Java/Others)    Memory Limit: 65535/65536 K (Java/Others)

Problem Description

Yuanfang is puzzled with the question below:

There are n integers, a₁, a₂, …, a_n. The initial values of them are 0. There are four kinds of operations.

Operation 1: Add c to each number between a_x and a_y inclusive. In other words, do transformation a_k<---a_k+c, k = x,x+1,…,y.

Operation 2: Multiply c to each number between a_x and a_y inclusive. In other words, do transformation a_k<---a_k×c, k = x,x+1,…,y.

Operation 3: Change the numbers between a_x and a_y to c, inclusive. In other words, do transformation a_k<---c, k = x,x+1,…,y.

Operation 4: Get the sum of p power among the numbers between a_x and a_y inclusive. In other words, get the result of a_x^p+a_x+1^p+…+a_y ^p.

Yuanfang has no idea of how to do it. So he wants to ask you to help him.

Input

There are no more than 10 test cases.

For each case, the first line contains two numbers n and m, meaning that there are n integers and m operations. 1 <= n, m <= 100,000.

Each the following m lines contains an operation. Operation 1 to 3 is in this format: "1 x y c" or "2 x y c" or "3 x y c". Operation 4 is in this format: "4 x y p". (1 <= x <= y <= n, 1 <= c <= 10,000, 1 <= p <= 3)

The input ends with 0 0.

Output

For each operation 4, output a single integer in one line representing the result. The answer may be quite large. You just need to calculate the remainder of the answer when divided by 10007.

Sample Input

5 5
3 3 5 7
1 2 4 4
4 1 5 2
2 2 5 8
4 3 5 3
0 0

Sample Output

307
7489

给你一个数组，初始值为零，有四种操作：（1）"1 x y c",代表 把区间 [x,y] 上的值全部加c。（2）"2 x y c",代表 把区间 [x,y] 上的值全部乘以c。（3）"3 x y c" 代表 把区间 [x,y]上的值全部赋值为c。

（4）"4 x y p" 代表 求区间 [x,y] 上值的p次方和1<=p<=3。

思路:

首先先解决一个区间内可能同时出现+和*的操作的问题(如果出现赋值,则在赋值之前的+和*都是无效的),我们可以把加法累计起来,把乘法先处理了,比如说原来[l,r]这一段上有了+x,现在要求*y,我们可以先把原来的每个数乘以y,然后加上x*y,即都是先处理乘法,在处理加法。

然后处理一下区间查询的问题,假设原来的值是k,如果是乘法或者赋值的话,sumv1,sumv2,sumv3的修改是很容易实现的,如果是加,(p+k)^3=p^3+3*p^2*k+3*k^2*p+k^3,所以sumv3[rt]=sumv3[rt]+sumv2[rt]*3*k+sumv1[rt]*3*k+(r-l+1)*k^3,同理二次方的修改也是如此

#include<bits/stdc++.h>
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
const int maxn=110100;
typedef long long ll;
const int MOD=10007;
ll sumv1[maxn*4],sumv2[maxn*4],sumv3[maxn*4],v,op,setv[maxn*4],mul[4*maxn],addv[4*maxn];
int L,R;

void build(int l,int r,int rt){
    sumv1[rt]=0,sumv2[rt]=0,sumv3[rt]=0;
    setv[rt]=addv[rt]=mul[rt]=0;
    if(l==r)
        return ;
    int mid=(l+r)>>1;
    build(lson);
    build(rson);
}

void pushup(int rt){
     sumv1[rt]=(sumv1[rt<<1]+sumv1[rt<<1|1])%MOD;
     sumv2[rt]=(sumv2[rt<<1]+sumv2[rt<<1|1])%MOD;
     sumv3[rt]=(sumv3[rt<<1]+sumv3[rt<<1|1])%MOD;
}

void change(int op,int rt,int l,int r,ll k){
    if(op==1){
        sumv3[rt]=(sumv3[rt]+sumv2[rt]*3*k+3*k*k*sumv1[rt]%MOD+k*k*k%MOD*(r-l+1))%MOD;
        sumv2[rt]=(sumv2[rt]+k*k*(r-l+1)+2*k*sumv1[rt])%MOD;
        sumv1[rt]=(sumv1[rt]+k*(r-l+1))%MOD;
    }
    else if(op==2){
        sumv3[rt]=sumv3[rt]*(k*k*k%MOD)%MOD;
        sumv2[rt]=sumv2[rt]*k*k%MOD;
        sumv1[rt]=sumv1[rt]*k%MOD;
    }
    else{
        sumv3[rt]=(k*k*k%MOD)*(r-l+1)%MOD;
        sumv2[rt]=k*k*(r-l+1)%MOD;
        sumv1[rt]=k*(r-l+1)%MOD;
    }
}

void pushdown(int rt,int l,int r){
    if(setv[rt]!=0){
        int mid=(l+r)>>1;
        change(3,rt<<1,l,mid,setv[rt]);
        change(3,rt<<1|1,mid+1,r,setv[rt]);
        setv[rt<<1]=setv[rt<<1|1]=setv[rt];
        addv[rt<<1]=addv[rt<<1|1]=0;
        mul[rt<<1]=mul[rt<<1|1]=0;
        setv[rt]=0;
    }
    if(mul[rt]!=0){
        int mid=(l+r)>>1;
        change(2,rt<<1,l,mid,mul[rt]);
        change(2,rt<<1|1,mid+1,r,mul[rt]);
        if(mul[rt<<1]==0)
            mul[rt<<1]=mul[rt];
        else
            mul[rt<<1]=mul[rt<<1]*mul[rt]%MOD;
        if(mul[rt<<1|1]==0)
            mul[rt<<1|1]=mul[rt];
        else
            mul[rt<<1|1]=mul[rt<<1|1]*mul[rt]%MOD;
        addv[rt<<1]=addv[rt<<1]*mul[rt]%MOD;
        addv[rt<<1|1]=addv[rt<<1|1]*mul[rt]%MOD;
        mul[rt]=0;
    }
    if(addv[rt]!=0){
        int mid=(l+r)>>1;
        change(1,rt<<1,l,mid,addv[rt]);
        change(1,rt<<1|1,mid+1,r,addv[rt]);
        addv[rt<<1]=(addv[rt<<1]+addv[rt])%MOD;
        addv[rt<<1|1]=(addv[rt<<1|1]+addv[rt])%MOD;
        addv[rt]=0;
    }
}

void update(int l,int r,int rt){
    if(L<=l&&R>=r){
        if(op==3){
            setv[rt]=v,mul[rt]=0,addv[rt]=0;
            change(op,rt,l,r,v);
        }
        else if(op==2){
            if(mul[rt]==0)
                mul[rt]=v;
            else
                mul[rt]=mul[rt]*v%MOD;
            addv[rt]=addv[rt]*v%MOD;
            change(op,rt,l,r,v);
        }
        else{
            addv[rt]=(addv[rt]+v)%MOD;
            change(op,rt,l,r,v);
        }
        return ;
    }
    pushdown(rt,l,r);
    int mid=(l+r)>>1;
    if(L<=mid)
        update(lson);
    if(R>mid)
        update(rson);
    pushup(rt);
}

ll query(int l,int r,int rt){
    if(L<=l&&R>=r){
        if(v==1)
            return sumv1[rt];
        if(v==2)
            return sumv2[rt];
        return sumv3[rt];
    }
    pushdown(rt,l,r);
    int mid=(l+r)>>1;
    ll ans=0;
    if(L<=mid)
        ans+=query(lson);
    if(R>mid)
        ans+=query(rson);
    return ans;
}

int main(){
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF){
        if(n==0&&m==0)
            break;
        build(1,n,1);
        for(int i=1;i<=m;i++){
            scanf("%d%d%d%lld",&op,&L,&R,&v);
            if(op==4)
                printf("%lld\n",query(1,n,1)%MOD);
            else
                update(1,n,1);
        }
    }
    return 0;
}