题意： 中文题
思路：
题目要求维护区间两两数的乘积，可以转化为维护区间的平方和。
需要用到逆元
// Decline is inevitable,
// Romance will last forever.
//#include <bits/stdc++.h>
#include <iostream>
#include <cmath>
#include <cstring>
#include <string>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <deque>
#include <vector>
using namespace std;
#define mst(a, x) memset(a, x, sizeof(a))
#define INF 0x3f3f3f3f
#define mp make_pair
#define pii pair<int,int>
#define fi first
#define se second
#define ll long long
#define int long long
const int maxn = 1e2 + 10;
const int maxm = 1e3 + 10;
//const int P = 1e4 + 7;
int P;
int a[maxn];
int n;
ll power(ll a, ll b) {
ll ans = 1 % P;
for (; b; b >>= 1) {
if (b & 1) ans = ans * a % P;
a = a * a % P;
}
return ans;
}
struct segement_tree {
int l, r;
int sum; //分别存储区间和
int sum2, sum3; //区间平方和，立方和
int add, mul, set; //lazy标记 分别对应加法，乘法，变成
#define sum(x) tree[x].sum
#define sum2(x) tree[x].sum2
#define sum3(x) tree[x].sum3
#define add(x) tree[x].add
#define mul(x) tree[x].mul
#define set(x) tree[x].set
#define l(x) tree[x].l
#define r(x) tree[x].r
#define ls p<<1, l, mid
#define rs p<<1, mid + 1, r
}tree[maxn << 2];
void push(int p) {
sum(p) = (sum(p<<1) + sum(p<<1|1)) % P; //不要忘记mod
sum2(p) = (sum2(p<<1) + sum2(p<<1|1)) % P;
}
void build(int p, int l, int r) {
l(p) = l; r(p) = r;
mul(p) = 1; //赋初始值
set(p) = add(p) = 0;
if(l == r) {
sum(p) = a[l] % P; //不要忘记mod
sum2(p) = a[l] * a[l] % P;
return;
}
int mid = (l + r) >> 1;
build(p<<1, l, mid);
build(p<<1|1, mid+1, r);
push(p);
}
void spread(int p) {
if(mul(p) != 1) {
mul(p<<1) = mul(p) * mul(p<<1) % P;
mul(p<<1|1) = mul(p) * mul(p<<1|1) % P;
if(add(p<<1))
add(p<<1) = add(p<<1) * mul(p) % P;
if(add(p<<1|1))
add(p<<1|1) = add(p<<1|1) * mul(p)%P;
sum(p<<1) = sum(p<<1) * mul(p) % P;
sum(p<<1|1) = sum(p<<1|1)*mul(p)%P;
sum2(p<<1) = sum2(p<<1)*mul(p)%P*mul(p)%P;
sum2(p<<1|1) = sum2(p<<1|1)*mul(p)%P*mul(p)%P;
mul(p) = 1;
}
if(add(p)) {
add(p<<1) = (add(p<<1)+add(p))%P;
add(p<<1|1) = (add(p<<1|1)+add(p))%P; //
ll lenl = r(p<<1)-l(p<<1)+1;
ll lenr = r(p<<1|1)-l(p<<1|1)+1;
sum2(p<<1)=(sum2(p<<1)+(add(p)*add(p)%P)*lenl%P+2*sum(p<<1)*add(p)%P)%P;
sum2(p<<1|1)=(sum2(p<<1|1)+(add(p)*add(p)%P)*lenr%P+2*sum(p<<1|1)*add(p)%P)%P;
sum(p<<1) = (sum(p<<1) + lenl * add(p)%P)% P;
sum(p<<1|1) = (sum(p<<1|1) + lenr * add(p)%P)% P;
add(p) = 0;
}
}
void mul_change(int p, int l, int r, int k) {
if(l <= l(p) && r >= r(p)) {
mul(p) = mul(p) * k % P;
sum(p) = sum(p) * k % P;
add(p) = add(p) * k % P;
sum2(p) = sum2(p) * k % P * k % P;
return;
}
spread(p);
int mid = (l(p) + r(p)) >> 1;
if(l <= mid) mul_change(p<<1, l, r, k);
if(r > mid) mul_change(p<<1|1, l, r, k);
push(p);
}
void add_change(int p, int l, int r, int d) {
if(l <= l(p) && r >= r(p)) {
add(p) = (add(p) + d) % P;
sum2(p)=(sum2(p)+(d*d%P* (r(p)-l(p)+1))%P + 2 * sum(p) * d) % P;
sum(p) = (sum(p) + (r(p)-l(p)+1)*d) % P;
return;
}
spread(p);
int mid = (l(p) + r(p)) >> 1;
if(l <= mid) add_change(p<<1, l, r, d);
if(r > mid) add_change(p<<1|1, l, r, d);
push(p);
}
void set_change(int p, int l, int r, int d) {
if(l <= l(p) && r >= r(p)) {
set(p) = d;
add(p) = 0; //清空标记
mul(p) = 1;
int len = r(p) - l(p) + 1;
sum(p) = d * len;
sum2(p) = len*d%P*d%P;
sum3(p) = len*d%P*d%P*d%P;
return;
}
spread(p);
int mid = (l(p) + r(p)) >> 1;
if(l <= mid)
set_change(p<<1, l, r, d);
if(r > mid)
set_change(p<<1|1, l, r, d);
push(p);
}
int query(int p, int l, int r, int index) { //index=1 2
if(l <= l(p) && r >= r(p)) {
if(index == 1)
return sum(p) % P;
else
return sum2(p) % P;
}
spread(p);
int mid = (l(p) + r(p)) >> 1;
int ans = 0;
if(l <= mid) ans += query(p<<1, l, r, index);
if(r > mid) ans += query(p<<1|1, l, r, index);
return ans % P; //不要忘记取mod
}
int m;
void solve() {
cin >> n >> m >> P;
bool ok = false;
if(P == 2) {
ok = true;
P = 3;
}
for(int i = 1; i <= n; i++)
{
cin >> a[i];
a[i] %= P;
}
build(1, 1, n);
while(m--) {
int ope, l, r, v;
cin >> ope >> l >> r;
if(ope == 3) {
ll ans1 = query(1, l, r, 1) * query(1, l, r, 1) % P;
ll ans2 = query(1, l, r, 2) % P;
ll ans;
if(ok) {
ans=((ans1-ans2)%P+P)%P;
if(ans)
ans=1;
}
else
ans=((ans1-ans2)%P+P)%P*power(2,P-2)%P;
cout << ans << endl;
}
else {
cin >> v;
v %= P;
if(ope == 1)
add_change(1, l, r, v);
else
mul_change(1, l, r, v);
}
}
}
signed main() {
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
// int T; scanf("%d", &T); while(T--)
// freopen("1.txt","r",stdin);
// freopen("output.txt","w",stdout);
int T; cin >> T;while(T--)
solve();
return 0;
}