树状数组

支持单点修改

#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std;

#define root 1, 1, N
#define ls o << 1, l, m
#define rs o << 1 | 1, m + 1, r

int L, R;
long long add[1500005], mul[1500005], c[1500005], P;

inline int read()
{
int x = 0, f = 1;
char ch = getchar();
while(!isdigit(ch))
{
if(ch == '-') f = -1;
ch = getchar();
}
while(isdigit(ch))
{
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}

inline void pushup(int o)
{
c[o] = (c[o << 1] + c[o << 1 | 1]) % P;
}

inline void build(int o, int l, int r)
{
add[o] = 0;
mul[o] = 1;
if(l == r)
{
scanf("%lld", &c[o]);
return;
}
int m = (l + r) >> 1;
build(ls);
build(rs);
pushup(o);
}

inline void pushdown(int o, int m)
{
if(add[o] == 0 && mul[o] == 1) return;
c[o << 1] = (c[o << 1] * mul[o] + add[o] * (m - (m >> 1))) % P;
c[o << 1 | 1] = (c[o << 1 | 1] * mul[o] + add[o] * (m >> 1)) % P;
add[o << 1] = (add[o << 1] * mul[o] + add[o]) % P;
add[o << 1 | 1] = (add[o << 1 | 1] * mul[o] + add[o]) % P;
mul[o << 1] = (mul[o << 1] * mul[o]) % P;
mul[o << 1 | 1] = (mul[o << 1 | 1] * mul[o]) % P;
add[o] = 0;
mul[o] = 1;
}

inline void update(int f, int d, int o, int l, int r)
{
if(L <= l && r <= R)
{
if(f == 2)
{
add[o] = (add[o] + d) % P;
c[o] = (c[o] + d * (r - l + 1)) % P;
}
else
{
mul[o] = (mul[o] * d) % P;
add[o] = (add[o] * d) % P;
c[o] = (c[o] * d) % P;
}
return;
}
pushdown(o, r - l + 1);
int m = (l + r) >> 1;
if(L <= m) update(f, d, ls);
if(m < R) update(f, d, rs);
pushup(o);
}

inline long long query(int o, int l, int r)
{
if(L <= l && r <= R) return c[o];
pushdown(o, r - l + 1);
int m = (l + r) >> 1;
long long ret = 0;
if(L <= m) ret += query(ls);
if(m < R) ret += query(rs);
return ret;
}

int main()
{
int N, Q;
N = read();
P = read();
build(root);
Q = read();
while(Q--)
{
int a, x, y, k;
a = read();
if(a == 1 || a == 2)
{
x = read();
y = read();
k = read();
L = x;
R = y;
update(a, k, root);
}
else
{
x = read();
y = read();
L = x;
R = y;
printf("%lld\n", query(root) % P);
}
}
return 0;
}

 ps:有意思的是这个代码还是[AHOI2009]维护序列的题解