这道题是一个典型的区间查询问题，具体代码如下（这是普通的写法，会TLE）
当然，这段代码运用了前缀和的思想，用空间换取了时间

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

constexpr ll MOD = 1e9 + 7;
int main() {
	int n, _;
	scanf("%d %d",&n,&_);
	vector<ll> a(n),b(n+10);
	for (auto &i:a)cin>>i;
    b[0] = a[0];
	for (int i = 1; i < n; i ++)
		b[i] = a[i] - a[i - 1];
	while (_ --) {
		int x; scanf("%d",&x);
		if (x == 1) {
			int l, r, k;
			scanf("%d %d %d",&l,&r,&k);
			b[l - 1] += k;
			b[r] -= k;
		}
		else {
			int l, r; scanf("%d %d",&l,&r);
			ll sum = 0;
			a[0] = b[0];
			for (int i = 1; i < n; i ++)
				a[i] = a[i - 1] + b[i];
			for (int i = l - 1; i < r; i ++) {
                ll h = a[i];
                h %= MOD;
                h *= a[i];
                h %= MOD;
                h *= a[i];
                h %= MOD;
				sum = (sum + h + MOD) % MOD;
            }
			printf("%d\n",sum);
		}
	}

	return 0;
}

所以我们需要用高级算法优化，这里要使用高级数据结构-线段树
具体代码如下

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAX=100005;
const ll MOD=1000000007;
ll sum1[4*MAX], sum2[4*MAX], sum3[4*MAX], lazyv[4*MAX];
ll A[MAX];
ll getMod(ll x){
    ll res=x%MOD;
    if(res<0) res+=MOD;
    return res;
}
void build(int node, int l, int r){
    if(l==r){
        ll val = getMod(A[l]);
        sum1[node]=val;
        sum2[node]=(val*val)%MOD;
        sum3[node]=(val*val%MOD*val)%MOD;
        return;
    }
    int mid=(l+r)/2;
    build(node*2,l,mid);
    build(node*2+1,mid+1,r);
    sum1[node]=(sum1[node*2]+sum1[node*2+1])%MOD;
    sum2[node]=(sum2[node*2]+sum2[node*2+1])%MOD;
    sum3[node]=(sum3[node*2]+sum3[node*2+1])%MOD;
}
void push_down(int node, int l, int r){
    if(lazyv[node]==0) return;
    ll K=lazyv[node];
    int mid=(l+r)/2, left=node*2, right=node*2+1, cnt1=mid-l+1, cnt2=r-mid;
    // Left child
    ll s1=sum1[left], s2=sum2[left];
    ll K2=(K*K)%MOD, K3=(K2*K)%MOD;
    sum1[left]=(sum1[left]+K*cnt1)%MOD;
    sum2[left]=(sum2[left]+(2*K*s1)%MOD + (K2*cnt1)%MOD)%MOD;
    sum3[left]=(sum3[left]+(3*K*s2)%MOD + (3*K2*s1)%MOD + (K3*cnt1)%MOD)%MOD;
    lazyv[left]=(lazyv[left]+K)%MOD;
    // Right child
    ll s1r=sum1[right], s2r=sum2[right];
    sum1[right]=(sum1[right]+K*cnt2)%MOD;
    sum2[right]=(sum2[right]+(2*K*s1r)%MOD + (K2*cnt2)%MOD)%MOD;
    sum3[right]=(sum3[right]+(3*K*s2r)%MOD + (3*K2*s1r)%MOD + (K3*cnt2)%MOD)%MOD;
    lazyv[right]=(lazyv[right]+K)%MOD;
    lazyv[node]=0;
}
void update(int node, int l, int r, int L, int R, ll K){
    if(R<l||r<L) return;
    if(L<=l && r<=R){
        ll cnt=r-l+1;
        ll s1=sum1[node], s2=sum2[node];
        ll K2=(K*K)%MOD, K3=(K2*K)%MOD;
        sum1[node]=(sum1[node]+K*cnt)%MOD;
        sum2[node]=(sum2[node]+(2*K*s1)%MOD + (K2*cnt)%MOD)%MOD;
        sum3[node]=(sum3[node]+(3*K*s2)%MOD + (3*K2*s1)%MOD + (K3*cnt)%MOD)%MOD;
        lazyv[node]=(lazyv[node]+K)%MOD;
        return;
    }
    push_down(node,l,r);
    int mid=(l+r)/2;
    update(node*2,l,mid,L,R,K);
    update(node*2+1,mid+1,r,L,R,K);
    sum1[node]=(sum1[node*2]+sum1[node*2+1])%MOD;
    sum2[node]=(sum2[node*2]+sum2[node*2+1])%MOD;
    sum3[node]=(sum3[node*2]+sum3[node*2+1])%MOD;
}
ll query(int node, int l, int r, int L, int R){
    if(R<l||r<L) return 0;
    if(L<=l && r<=R) return sum3[node];
    push_down(node,l,r);
    int mid=(l+r)/2;
    return (query(node*2,l,mid,L,R)+query(node*2+1,mid+1,r,L,R))%MOD;
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int N,M; cin>>N>>M;
    for(int i=1;i<=N;i++) cin>>A[i];
    build(1,1,N);
    while(M--){
        int type; cin>>type;
        if(type==1){
            int L,R; ll K; cin>>L>>R>>K;
            K=getMod(K);
            update(1,1,N,L,R,K);
        }
        else{
            int L,R; cin>>L>>R;
            ll ans=query(1,1,N,L,R);
            ans=getMod(ans);
            cout<<ans<<"\n";
        }
    }
    return 0;
}

希望大家多多AC哦，希望这个题解能帮到你

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