分块新技能 $+1$。

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10,M=1010,P=10007;
int a[N],tag1[M],tag2[M],n,B;
void pushdown(int x)
{
	int bl=(x-1)*B+1,br=min(n,x*B);
	for(int i=bl;i<=br;i++)a[i]=(a[i]*tag2[x]+tag1[x])%P;
	tag1[x]=0,tag2[x]=1;
}
void upd1(int l,int r,int x)
{
	int bl=(l-1)/B+1,br=(r-1)/B+1;
	if(bl==br)
	{
		pushdown(bl);
		for(int i=l;i<=r;i++)
			a[i]=(a[i]+x)%P;
	}
	else
	{
		pushdown(bl);pushdown(br);
		for(int i=l;i<=bl*B;i++)a[i]=(a[i]+x)%P;
		for(int i=(br-1)*B+1;i<=r;i++)a[i]=(a[i]+x)%P;
		for(int i=bl+1;i<br;i++)tag1[i]=(tag1[i]+x)%P;
	} 
}
void upd2(int l,int r,int x)
{
	int bl=(l-1)/B+1,br=(r-1)/B+1;
	if(bl==br)
	{
		pushdown(bl);
		for(int i=l;i<=r;i++)
			a[i]=a[i]*x%P;
	}
	else
	{
		pushdown(bl);pushdown(br);
		for(int i=l;i<=bl*B;i++)a[i]=a[i]*x%P;
		for(int i=(br-1)*B+1;i<=r;i++)a[i]=a[i]*x%P;
		for(int i=bl+1;i<br;i++)
		{
			tag1[i]=tag1[i]*x%P;
			tag2[i]=tag2[i]*x%P;
		}
	}
}
signed main()
{
	cin>>n;B=sqrt(n);
	for(int i=1;i<=n;i++)cin>>a[i];
	for(int i=1;i<=(n-1)/B+1;i++)tag2[i]=1;
	for(int i=1;i<=n;i++)
	{
		int op,l,r,c;cin>>op>>l>>r>>c;
		if(op==0)upd1(l,r,c);
		if(op==1)upd2(l,r,c);
		if(op==2)pushdown((r-1)/B+1),cout<<a[r]<<"\n";
	}
	return 0;
}

#include <bits/stdc++.h>
#define int long long
using namespace std;

const int N = 1e5 + 10, sqrtN = 350, MOD = 10007;
// 区间加法，区间乘法，单点查询
// a[i]记录每个点的值，b[i]记录每个点i所在块;
// add_tag[i]记录第i个块的加法标记，mul_tag[i]记录第i个块的乘法标记
// 每个块i的左端点L[i]、右端点R[i]
int n, a[N], b[N], L[sqrtN], R[sqrtN], add_tag[sqrtN], mul_tag[sqrtN];

void push_down(int x) { // 下传第x个块的标记
    for (int i = L[x]; i <= R[x]; i++) {
        a[i] = (a[i] * mul_tag[x] + add_tag[x]) % MOD;
    }
    mul_tag[x] = 1;
    add_tag[x] = 0;
}

signed main() {
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        a[i] %= MOD;
    }
    int B = sqrt(n), cnt = (n + B - 1) / B; // B为每块的长度，cnt为总块数
    for (int i = 1; i <= n; i++) {
        b[i] = (i - 1) / B + 1;
    }
    for (int i = 1; i <= cnt; i++) {
        L[i] = (i - 1) * B + 1;
        R[i] = min(i * B, n);
    }
    memset(add_tag, 0, sizeof(add_tag));
    for (int i = 1; i <= cnt; i++) mul_tag[i] = 1;
    for (int i = 1; i <= n; i++) {
        int op, l, r, c;
        cin >> op >> l >> r >> c;
        if (op == 0) { // 区间加
            if (b[l] == b[r]) { // 如果l和r在同一块内
                push_down(b[l]);
                for (int j = l; j <= r; j++)
                    a[j] = (a[j] + c) % MOD;
            } else {
                push_down(b[l]);
                for (int j = l; j <= R[b[l]]; j++)
                    a[j] = (a[j] + c) % MOD;
                for (int j = b[l] + 1; j <= b[r] - 1; j++)
                    add_tag[j] = (add_tag[j] + c) % MOD;
                push_down(b[r]);
                for (int j = L[b[r]]; j <= r; j++)
                    a[j] = (a[j] + c) % MOD;
            }
        } else if (op == 1) { // 区间乘
            if (b[l] == b[r]) { // 如果l和r在同一块内
                push_down(b[l]);
                for (int j = l; j <= r; j++)
                    a[j] = (a[j] * c) % MOD;
            } else {
                push_down(b[l]);
                for (int j = l; j <= R[b[l]]; j++)
                    a[j] = (a[j] * c) % MOD;
                for (int j = b[l] + 1; j <= b[r] - 1; j++) {
                    mul_tag[j] = (mul_tag[j] * c) % MOD;
                    add_tag[j] = (add_tag[j] * c) % MOD;
                }
                push_down(b[r]);
                for (int j = L[b[r]]; j <= r; j++)
                    a[j] = (a[j] * c) % MOD;
            }
        } else { // 单点查询
            cout << (a[r] * mul_tag[b[r]] + add_tag[b[r]]) % MOD << '\n';
        }
    }
    return 0;
}