题目：Transformation
这是我参考的博客：https://blog.csdn.net/a664607530/article/details/74563018

总结：这题是看了将近一个上午才看明白，这里面线段树的节点保存的是这个节点所在区间内相同的元素的值，比如1~mid，这个节点的值为3，说明的是1 ~mid,所有元素的值都是为3
if(lazy[rt]){
    lazy[ls] = lazy[rs] = true;
    lazy[rt] = false;
    tree[ls] = tree[rs] = tree[rt];
}

这段代码的意思是住下继续扩展，因为当前区间不满足操作
void pushup(int rt){
    if(lazy[rs] == false || lazy[ls] == false){
        lazy[rt] = false;
    }else if(tree[ls] != tree[rs]){
        lazy[rt] = false;
    }else {
        lazy[rt] = true;
        tree[rt] = tree[ls];
    }
}

这段代码就有点类似于合并了，根据下面两个子区间的情况，判断父亲区间是否要发生变化。
#include <stdio.h>
#include <string.h>
#include <algorithm>
#define mid (start+end)/2
#define ls (2*rt)
#define rs ((2*rt)+1)
const int mod = 10007;
const int N = 1e5+5;
using namespace std;

int tree[N<<2];
bool lazy[N<<2];
int n,m;

void pushup(int rt){
    if(lazy[rs] == false || lazy[ls] == false){
        lazy[rt] = false;
    }else if(tree[ls] != tree[rs]){
        lazy[rt] = false;
    }else {
        lazy[rt] = true;
        tree[rt] = tree[ls];
    }
}
void update(int rt,int start,int end,int l,int r,int v,int op){
    if(lazy[rt] && start >= l && end <= r){
        if(op == 1){
            tree[rt] = (tree[rt]+v)%mod;
        }else if(op == 2){
            tree[rt] = (tree[rt]*v)%mod;
        }else {
            tree[rt] = v%mod;
        }
        return;
    }
    if(lazy[rt]){
        lazy[ls] = lazy[rt] = true;
        tree[ls] = tree[rs] = tree[rt];
        lazy[rt] = false;
    }
    if(l <= mid) update(ls,start,mid,l,r,v,op);
    if(r > mid) update(rs,mid+1,end,l,r,v,op);
    pushup(rt);
}
int query(int rt,int start,int end,int l,int r,int p){
    if(lazy[rt] && start >= l && end <= r){
        int ans = 1;
        for(int i = 1;i <= p;i++){
            ans = (ans*tree[rt])%mod;
        }
        ans = (ans*(end-start+1))%mod;
        return ans;
    }
    if(lazy[rt]){
        lazy[ls] = lazy[rs] = true;
        lazy[rt] = false;
        tree[ls] = tree[rs] = tree[rt];
    }
    int ans = 0;
    if(l <= mid) ans += query(ls,start,mid,l,r,p);
    if(r > mid) ans += query(rs,mid+1,end,l,r,p);
    return ans%mod;
}
void traver(){
    for(int i = 1;i <= 9;i++){
        printf("%3d",i);
    }
    printf("\n");
    for(int i = 1;i <= 9;i++){
        printf("%3d",lazy[i]);
    }
    printf("\n");
    for(int i = 1;i <= 9;i++){
        printf("%3d",tree[i]);
    }
    printf("\n\n");
}
int main(){
    while(scanf("%d%d",&n,&m)&&(n+m)){
        memset(tree,0,sizeof tree);
        memset(lazy,true,sizeof lazy);
        while(m--){
            int op,l,r,p;
            scanf("%d%d%d%d",&op,&l,&r,&p);
            if(op >= 1 && op <= 3){
                update(1,1,n,l,r,p,op);
            }else {
                printf("%d\n",query(1,1,n,l,r,p));
            }
        //    traver();
        }
    }
    return 0;
}