先套用一个线段树维护离散化之后的区间的每一段的答案
那么只要考虑怎么下面的东西即可
\[\sum_{i=1}^{n}(A\times i \ mod \ B)
\]
\]
拆开就是
\[\sum_{i=1}^{n}A\times i-B\times \sum_{i=1}^{n}\lfloor\frac{A\times i}{B}\rfloor
\]
\]
只要考虑计算 \(\sum_{i=1}^{n}\lfloor\frac{A\times i}{B}\rfloor\) 即可
类欧几里德算法
若 \(A>B\)，那么就是
\[\lfloor\frac{A}{B}\rfloor\sum_{i=1}^{n}i+\sum_{i=1}^{n}\lfloor\frac{(A \ mod \ B)\times i}{B}\rfloor
\]
\]
变成 \(A<B\) 的情况
对于 \(A<B\)，那么可以看成是求直线 \(y=\frac{A}{B}\times i,i\in[1,n]\) 与坐标轴围成的三角形中的整点的个数
设 \(m=\lfloor\frac{A\times n}{B}\rfloor\)
把问题为矩形 \((0,0),(n,m)\) 内的减去三角形 \((0,0),(n,m),(0,m)\) 内的再加上对角线的
对角线上的就是 \(\frac{n\times gcd(A,B)}{B}\)，矩形内的就是 \(n\times m\)
对于那个三角形的，反转坐标系后相当于是求直线 \(y=\frac{B}{A}\times i,i\in[1,\lfloor\frac{A\times n}{B}\rfloor]\) 与坐标轴围成的三角形中的整点的个数
即
\[\sum_{i=1}^{\lfloor\frac{A\times n}{B}\rfloor}\lfloor\frac{B\times i}{A}\rfloor
\]
\]
递归处理即可，复杂度和 \(gcd\) 一样
可以先把 \(A,B\) 同时除去 \(gcd\) 后再做
这个题注意一个细节
\[\sum_{i=1}^{n}A\times i-B\times \sum_{i=1}^{n}\lfloor\frac{A\times i}{B}\rfloor
\]
\]
直接求的可能会爆 \(long \ long\)
可以对 \(B\) 分段，算出长度为 \(B\) 的乘上 \(\lfloor\frac{n}{B}\rfloor\)，再加上长度为 \(n \ mod \ B\) 的，可以接受
# include <bits/stdc++.h>
using namespace std;
typedef long long ll;
namespace IO {
const int maxn(1 << 21 | 1);
char ibuf[maxn], *iS, *iT, c;
int f;
inline char Getc() {
return iS == iT ? (iT = (iS = ibuf) + fread(ibuf, 1, maxn, stdin), (iS == iT ? EOF : *iS++)) : *iS++;
}
template <class Int> inline void In(Int &x) {
for (f = 1, c = Getc(); c < '0' || c > '9'; c = Getc()) f = c == '-' ? -1 : 1;
for (x = 0; c >= '0' && c <= '9'; c = Getc()) x = (x << 1) + (x << 3) + (c ^ 48);
x *= f;
}
}
using IO :: In;
const int maxn(1e5 + 5);
int n, m, o[maxn], cnt;
struct Segment {
ll sum;
int a, b, l;
} tr[maxn << 2];
inline void Update(int x) {
tr[x].sum = tr[x << 1].sum + tr[x << 1 | 1].sum;
}
inline int Gcd(int a, int b) {
return !b ? a : Gcd(b, a % b);
}
inline ll Mul(int a, int b, int len) {
register int k = a / b;
register ll sum = 1LL * k * (len + 1) * len / 2;
if (!(a %= b) || !b) return sum;
register int m = 1LL * len * a / b;
assert(m >= 0);
return 1LL * len * m + len / b - Mul(b, a, m) + sum;
}
inline ll Calc(int a, int b, int len) {
if (len < 1 || a == b || b == 1) return 0;
register int g = Gcd(a, b);
return 1LL * a * len * (len + 1) / 2 - 1LL * b * Mul(a / g, b / g, len);
}
inline ll Solve(int a, int b, int len) {
register ll ret = Calc(a, b, len % b);
if (len >= b) ret += 1LL * len / b * Calc(a, b, b);
return ret;
}
inline void Add(int x, int a, int b, int l, int len) {
tr[x].a = a, tr[x].b = b, tr[x].l = l;
tr[x].sum = Solve(a, b, l + len - 1) - Solve(a, b, l - 1);
}
inline void Pushdown(int x, int l, int r) {
if (!tr[x].a) return;
register int mid = (l + r) >> 1;
Add(x << 1, tr[x].a, tr[x].b, tr[x].l, o[mid] - o[l]);
Add(x << 1 | 1, tr[x].a, tr[x].b, tr[x].l + o[mid] - o[l], o[r] - o[mid]);
tr[x].a = tr[x].b = tr[x].l = 0;
}
ll Query(int x, int l, int r, int ql, int qr) {
if (l >= qr || r <= ql) return 0;
if (ql <= l && qr >= r) return tr[x].sum;
Pushdown(x, l, r);
register int mid = (l + r) >> 1;
register ll ret = 0;
if (ql <= mid) ret = Query(x << 1, l, mid, ql, qr);
if (qr >= mid) ret += Query(x << 1 | 1, mid, r, ql, qr);
return ret;
}
void Modify(int x, int l, int r, int ql, int qr, int a, int b) {
if (l >= qr || r <= ql) return;
if (ql <= l && qr >= r) {
Add(x, a, b, o[l] - o[ql] + 1, o[r] - o[l]);
return;
}
Pushdown(x, l, r);
register int mid = (l + r) >> 1;
if (ql <= mid) Modify(x << 1, l, mid, ql, qr, a, b);
if (qr >= mid) Modify(x << 1 | 1, mid, r, ql, qr, a, b);
Update(x);
}
int op[maxn], ql[maxn], qr[maxn], a[maxn], b[maxn];
int main () {
In(n), In(m);
register int i, l, r;
for (i = 1; i <= m; ++i) {
In(op[i]), In(ql[i]), In(qr[i]), --ql[i];
o[++cnt] = ql[i], o[++cnt] = qr[i];
if (op[i] == 1) In(a[i]), In(b[i]);
}
sort(o + 1, o + cnt + 1), cnt = unique(o + 1, o + cnt + 1) - o - 1;
for (i = 1; i <= m; ++i) {
l = lower_bound(o + 1, o + cnt + 1, ql[i]) - o;
r = lower_bound(o + 1, o + cnt + 1, qr[i]) - o;
if (op[i] == 1) Modify(1, 1, cnt, l, r, a[i], b[i]);
else printf("%lld\n", Query(1, 1, cnt, l, r));
}
return 0;
}