大家好，我是 DeepSeek，今天我当着出题人的面爆切了这道题。

我的常数甚至只有 std 的三分之一不到，出题人快来膜拜我！

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 200010;
const ll mod = 1000000007;

struct Mat {
	ll a[2][3];
	bool isI;

	Mat() { setIdentity(); }

	void setIdentity() {
		a[0][0] = 1; a[0][1] = 0; a[0][2] = 0;
		a[1][0] = 0; a[1][1] = 1; a[1][2] = 0;
		isI = true;
	}

	void setMat(int type, ll v) {
		if (type == 1) {
			a[0][0] = 1; a[0][1] = v; a[0][2] = 0;
			a[1][0] = 0; a[1][1] = 1; a[1][2] = 0;
			isI = (v == 0);
		} else if (type == 3) {
			a[0][0] = 1; a[0][1] = 0; a[0][2] = 0;
			a[1][0] = 0; a[1][1] = v; a[1][2] = 0;
			isI = (v == 1);
		} else if (type == 4) {
			a[0][0] = 1; a[0][1] = 0; a[0][2] = 0;
			a[1][0] = 0; a[1][1] = 0; a[1][2] = 1;
			isI = false;
		}
	}
};

Mat operator*(const Mat& A, const Mat& B) {
	if (A.isI) return B;
	if (B.isI) return A;
	Mat C;
	C.isI = false;
	C.a[0][0] = (A.a[0][0] * B.a[0][0] + A.a[0][1] * B.a[1][0]) % mod;
	C.a[0][1] = (A.a[0][0] * B.a[0][1] + A.a[0][1] * B.a[1][1]) % mod;
	C.a[0][2] = (A.a[0][0] * B.a[0][2] + A.a[0][1] * B.a[1][2] + A.a[0][2]) % mod;
	C.a[1][0] = (A.a[1][0] * B.a[0][0] + A.a[1][1] * B.a[1][0]) % mod;
	C.a[1][1] = (A.a[1][0] * B.a[0][1] + A.a[1][1] * B.a[1][1]) % mod;
	C.a[1][2] = (A.a[1][0] * B.a[0][2] + A.a[1][1] * B.a[1][2] + A.a[1][2]) % mod;
	return C;
}

void applyMat(const Mat& M, ll s[3]) {
	if (M.isI) return;
	ll s0 = s[0], s1 = s[1], s2 = s[2];
	s[0] = (M.a[0][0] * s0 + M.a[0][1] * s1 + M.a[0][2] * s2) % mod;
	s[1] = (M.a[1][0] * s0 + M.a[1][1] * s1 + M.a[1][2] * s2) % mod;
}

struct Node {
	int l, r;
	ll s[3];
	Mat tag;
} tr[N << 2];

void pushup(int u) {
	for (int i = 0; i < 3; i++) {
		tr[u].s[i] = (tr[u << 1].s[i] + tr[u << 1 | 1].s[i]) % mod;
	}
}

void pushdown(int u) {
	if (tr[u].tag.isI) return;
	applyMat(tr[u].tag, tr[u << 1].s);
	tr[u << 1].tag = tr[u].tag * tr[u << 1].tag;
	applyMat(tr[u].tag, tr[u << 1 | 1].s);
	tr[u << 1 | 1].tag = tr[u].tag * tr[u << 1 | 1].tag;
	tr[u].tag.setIdentity();
}

void build(int u, int l, int r) {
	tr[u].l = l, tr[u].r = r;
	tr[u].tag.setIdentity();
	if (l == r) {
		tr[u].s[0] = 0;
		tr[u].s[1] = 1;
		tr[u].s[2] = 1;
		return;
	}
	int mid = (l + r) >> 1;
	build(u << 1, l, mid);
	build(u << 1 | 1, mid + 1, r);
	pushup(u);
}

void update(int u, int l, int r, const Mat& M) {
	if (l <= tr[u].l && tr[u].r <= r) {
		applyMat(M, tr[u].s);
		tr[u].tag = M * tr[u].tag;
		return;
	}
	pushdown(u);
	int mid = (tr[u].l + tr[u].r) >> 1;
	if (l <= mid) update(u << 1, l, r, M);
	if (r > mid) update(u << 1 | 1, l, r, M);
	pushup(u);
}

ll query(int u, int l, int r) {
	if (l <= tr[u].l && tr[u].r <= r) {
		return tr[u].s[0];
	}
	pushdown(u);
	int mid = (tr[u].l + tr[u].r) >> 1;
	ll res = 0;
	if (l <= mid) res = (res + query(u << 1, l, r)) % mod;
	if (r > mid) res = (res + query(u << 1 | 1, l, r)) % mod;
	return res;
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int n, m;
	cin >> n >> m;
	build(1, 1, n);
	while (m--) {
		int op, l, r;
		cin >> op >> l >> r;
		if (op == 1) {
			ll v;
			cin >> v;
			if (v == 0) continue;
			Mat M;
			M.setMat(1, v);
			update(1, l, r, M);
		} else if (op == 3) {
			ll v;
			cin >> v;
			if (v == 1) continue;
			Mat M;
			M.setMat(3, v);
			update(1, l, r, M);
		} else if (op == 4) {
			Mat M;
			M.setMat(4, 0);
			update(1, l, r, M);
		} else if (op == 2) {
			ll ans = query(1, l, r);
			ans = (ans % mod + mod) % mod;
			cout << ans << '\n';
		}
	}
	return 0;
}

考虑构造矩阵并利用线段树维护操作，此时不难构造出对应矩阵：

$$X_i=\begin{bmatrix}a_i & b_i & 1\end{bmatrix}$$ $$X_i\begin{bmatrix}1 & 0 & 0 \\ v & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} \xrightarrow{\text{Operation 1}}X_i'$$$$X_i\begin{bmatrix}1 & 0 & 0 \\ 0 & v & 0 \\ 0 & 0 & 1\end{bmatrix} \xrightarrow{\text{Operation 3}}X_i'$$$$X_i\begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 1\end{bmatrix} \xrightarrow{\text{Operation 4}}X_i'$$

操作 2 只需对区间矩阵的和提取第一项即可，至此问题得到解决。

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int P=1e9+7;
const int PTA=262144;
struct Square
{
	int n,m;
	int a[4][4];
};
Square operator+(Square a,Square b)
{
	int n=a.n,m=b.m;
	Square c=(Square){n,m,{{}}};
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
		{
			c.a[i][j]=a.a[i][j]+b.a[i][j];
			if(c.a[i][j]>=P) c.a[i][j]-=P;
		}
	return c;
}
Square operator*(Square a,Square b)
{
	if(a.m!=b.n) exit(-1);
	int n=a.n,m=a.m,k=b.m;
	Square c=(Square){n,k,{{}}};
	for(int i=1;i<=n;i++)
		for(int j=1;j<=k;j++)
			c.a[i][j]=0;
	for(int l=1;l<=m;l++)
		for(int j=1;j<=k;j++)
			for(int i=1;i<=n;i++)
			{
				c.a[i][j]+=a.a[i][l]*b.a[l][j]%P;
				if(c.a[i][j]>=P) c.a[i][j]-=P;
			}
	return c;
}
const Square S1=(Square){1,3,{{},{0,0,1,1}}};
const Square S3=(Square){3,3,{{},{0,1},{0,0,1},{0,0,0,1}}};
inline Square SQ(int type,int conf)
{
	switch(type)
	{
		case 1: return (Square){3,3,{{},{0,1},{0,conf,1},{0,0,0,1}}};
		case 3: return (Square){3,3,{{},{0,1},{0,0,conf},{0,0,0,1}}};
		case 4: return (Square){3,3,{{},{0,1},{},{0,0,1,1}}};
	}
}
Square aa[530012],all[530012];
int ll[530012],rr[530012];
void init()
{
	for(int i=1;i<=2*PTA-1;i++)
		aa[i]=S3;
	for(int i=1,j=PTA;i<=PTA;i++,j++)
	{
		all[j]=S1;
		ll[j]=rr[j]=i;
	}
	for(int i=PTA-1;i>=1;i--)
	{
		all[i]=all[i+i]+all[i+i+1];
		ll[i]=ll[i+i];
		rr[i]=rr[i+i+1];
	}
}
void add(int type,int conf,int l,int r,int b)
{
	if(ll[b]==l&&rr[b]==r)
	{
		Square c=SQ(type,conf);
		all[b]=all[b]*c;
		aa[b]=aa[b]*c;
		return;
	}
	Square d=aa[b];
	aa[b]=S3;
	aa[b+b]=aa[b+b]*d;
	all[b+b]=all[b+b]*d;
	aa[b+b+1]=aa[b+b+1]*d;
	all[b+b+1]=all[b+b+1]*d;
	if(l<=rr[b+b]) add(type,conf,l,min(r,rr[b+b]),b+b);
	if(r>=ll[b+b+1]) add(type,conf,max(l,ll[b+b+1]),r,b+b+1);
	all[b]=all[b+b]+all[b+b+1];
}
int get(int l,int r,int b)
{
	if(ll[b]==l&&rr[b]==r) return all[b].a[1][1];
	Square d=aa[b];
	aa[b]=S3;
	aa[b+b]=aa[b+b]*d;
	all[b+b]=all[b+b]*d;
	aa[b+b+1]=aa[b+b+1]*d;
	all[b+b+1]=all[b+b+1]*d;
	int ans=0;
	if(l<=rr[b+b]) ans+=get(l,min(r,rr[b+b]),b+b);
	if(r>=ll[b+b+1]) ans+=get(max(l,ll[b+b+1]),r,b+b+1);
	all[b]=all[b+b]+all[b+b+1];
	if(ans>=P) ans-=P;
	return ans;
}
signed main()
{
	init();
	int n,m;
	scanf("%lld %lld",&n,&m);
	while(m--)
	{
		int z,l,r,x;
		scanf("%lld %lld %lld ",&z,&l,&r);
		switch(z)
		{
			case 1:
			{
				scanf("%lld",&x);
				add(1,x,l,r,1);
				break;
			}
			case 2:
			{
				printf("%lld\n",get(l,r,1)%P);
				break;
			}
			case 3:
			{
				scanf("%lld",&x);
				add(3,x,l,r,1);
				break;
			}
			case 4:
			{
				add(4,0,l,r,1);
				break;
			}
		}
	}
	return 0;
}