codeforce题目链接：
CodeForces 1109E E. Sasha and a Very Easy Test
题解原发于我的blog
这是本蒟蒻发的第三篇黑题的题解，很开心。
一道很毒瘤的线段树题
本来蒟蒻兴高采烈的刷线段树的题，看到这道黑题，觉得很水决定刚一刚
这不是一道简单的线段树题
考虑直接相除，
结果(WA)了，这正是此题的难处
考虑使用乘法逆元
但是样例二直接否定了这种情况，除数可能不与模数互质。
考虑将模数分解质因数。
在懒标记中标记模数的每个质因子个数
相乘时相加质因子，相除时相减质因子
乘数中除模数质因子以外的数得累乘标记
除数除模数质因子以外的数可以直接乘法逆元计算
记住得使用欧拉定理不能使用费马小定理，因为模数不一定是质数
费马小定理对于乘法逆元推导(a和p互质，p为质数):
[a^{p-2}equiv a^{-1}(mod p)
]
欧拉定理对于乘法逆元推导(a和p互质):
[a^{varphi(p)-1}equiv a^{-1}(mod p)
]
其中
[varphi(x)=prod_{i=1}^{n}frac{p_i-1}{p_i}
]
(p_1, p_2……p_n)为(x)的所有质因数，(x)是不为(0)的整数。
最终在给几个标记的性质
- 模数不会超过有(9)个不同的质因子（把前九个质数累乘试一试）。
- 模数的单个质因子不会超过(20)个（(2^{20}>10^5)）
- 除模数质因子外的数可以直接取模（反正乘法逆元行得通）
假黑题
直接运算即可，一下是蒟蒻的评测记录
这么简单的代码我都调了这么久一定是我太菜了
最后上代码：
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cstring>
#include <cctype>
using namespace std;
template <typename T>
inline void read(T &x)
{
x = 0;
char s = getchar();
bool f = false;
while (!(s >= '0' && s <= '9'))
{
if (s == '-')
f = true;
s = getchar();
}
while (s >= '0' && s <= '9')
{
x = (x << 1) + (x << 3) + s - '0';
s = getchar();
}
if (f)
x = (~x) + 1;
}
namespace OUT
{
char Out[1 << 25], *fe = Out, ch[25];
int num = 0;
template <typename T>
inline void write(T x)
{
if (!x)
*fe++ = '0';
if (x < 0)
{
*fe++ = '-';
x = -x;
}
while (x)
{
ch[++num] = x % 10 + '0';
x /= 10;
}
while (num) *fe++ = ch[num--];
*fe++ = '
';
}
inline void flush()
{
fwrite(Out, 1, fe - Out, stdout);
fe = Out;
}
} // namespace OUT
using namespace OUT;
#define re register
#define ll long long
#define ls (k << 1)
#define rs (k << 1 | 1)
#define node vector<int>
const int N = 1e5 + 10, S = 40;
int n, q, mod, phi, pcnt, a[N];
vector<int> pm;
struct Tree
{
int c[S], sum, mul, high; // high是剩余的数
} tree[N << 2];
inline node dec(int n) //质因数分解
{
node res;
res.clear();
for (re int i = 2; i * i <= n; i++)
while (n % i == 0)
{
n /= i;
res.push_back(i);
}
if (n > 1)
res.push_back(n);
return res;
}
inline void Unique(node &res) //质因数去重(血的教训)
{
sort(res.begin(), res.end());
int u = unique(res.begin(), res.end()) - res.begin();
while (res.size() > u) res.pop_back();
}
inline int getphi(int n) //欧拉函数
{
int ans = n;
for (re int i = 2; i * i <= n; i++)
if (n % i == 0)
{
ans = ans / i * (i - 1);
while (n % i == 0) n /= i;
}
if (n > 1)
ans = ans / n * (n - 1);
return ans;
}
inline int pow(int a, int b, int mod = ::mod)
{
int ans = 1;
for (; b; b >>= 1, a = (ll)a * a % mod)
if (b & 1)
ans = (ll)ans * a % mod;
return ans;
}
inline int inv(int x)
{
return pow(x, phi - 1);
}
inline void get(int res, int &ans, int *c)
{
for (re int i = 1; i <= pcnt; i++)
while (res % pm[i - 1] == 0)
{
res /= pm[i - 1];
c[i]++;
}
ans = res % mod;
}
inline int calc(int v, int *dev)
{
for (re int i = 1; i <= pcnt; i++) v = (ll)v * pow(pm[i - 1], dev[i]) % mod;
return v;
}
inline void pushdown(int k)
{
for (re int i = 1; i <= pcnt; i++)
{
tree[ls].c[i] += tree[k].c[i];
tree[rs].c[i] += tree[k].c[i];
tree[k].c[i] = 0;
}
tree[ls].sum = (ll)tree[ls].sum * tree[k].mul % mod;
tree[rs].sum = (ll)tree[rs].sum * tree[k].mul % mod;
tree[ls].mul = (ll)tree[ls].mul * tree[k].mul % mod;
tree[rs].mul = (ll)tree[rs].mul * tree[k].mul % mod;
tree[ls].high = (ll)tree[ls].high * tree[k].high % mod;
tree[rs].high = (ll)tree[rs].high * tree[k].high % mod;
tree[k].mul = tree[k].high = 1;
}
inline void pushup(int k)
{
tree[k].sum = (tree[ls].sum + tree[rs].sum) % mod;
}
inline void build(int k, int l, int r)
{
tree[k].mul = tree[k].high = 1;
memset(tree[k].c, 0, sizeof(tree[k].c));
if (l == r)
{
tree[k].sum = a[l] % mod;
get(a[l], tree[k].high, tree[k].c);
return;
}
int mid = l + r >> 1;
build(ls, l, mid);
build(rs, mid + 1, r);
pushup(k);
}
inline void update1(int k, int l, int r, int x, int y, int v, int *dev, int high)
{
if (r < x || l > y)
return;
if (x <= l && r <= y)
{
for (re int i = 1; i <= pcnt; i++) tree[k].c[i] += dev[i];
tree[k].sum = (ll)tree[k].sum * v % mod;
tree[k].mul = (ll)tree[k].mul * v % mod;
tree[k].high = (ll)tree[k].high * high % mod;
return;
}
pushdown(k);
int mid = l + r >> 1;
update1(ls, l, mid, x, y, v, dev, high);
update1(rs, mid + 1, r, x, y, v, dev, high);
pushup(k);
}
inline void update2(int k, int l, int r, int pos, int v, int *dev, int high)
{
if (l == r)
{
tree[k].high = (ll)tree[k].high * inv(high) % mod;
for (re int i = 1; i <= pcnt; i++) tree[k].c[i] -= dev[i];
tree[k].sum = calc(tree[k].high, tree[k].c);
return;
}
pushdown(k);
int mid = l + r >> 1;
if (pos <= mid)
update2(ls, l, mid, pos, v, dev, high);
else
update2(rs, mid + 1, r, pos, v, dev, high);
pushup(k);
}
inline int query(int k, int l, int r, int x, int y)
{
if (r < x || l > y)
return 0;
if (x <= l && r <= y)
return tree[k].sum;
pushdown(k);
int mid = l + r >> 1;
return (query(ls, l, mid, x, y) + query(rs, mid + 1, r, x, y)) % mod;
}
int main()
{
read(n), read(mod), phi = getphi(mod);
for (re int i = 1; i <= n; i++) read(a[i]);
pm = dec(mod);
Unique(pm);
pcnt = pm.size();
build(1, 1, n);
read(q);
for (re int i = 1, c[S], opt, a, b, x, v; i <= q; i++)
{
read(opt), read(a), read(b);
memset(c, 0, sizeof(c));
if (opt == 1)
{
read(x);
get(x, v, c);
update1(1, 1, n, a, b, x, c, v);
}
else if (opt == 2)
{
get(b, v, c);
update2(1, 1, n, a, b, c, v);
}
else if (opt == 3)
write(query(1, 1, n, a, b));
}
flush();
return 0;
}
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