Luogu4433：[COCI2009-2010#1] ALADIN(类欧几里德算法)

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 imes i mod B) ]

拆开就是

[sum_{i=1}^{n}A imes i-B imes sum_{i=1}^{n}lfloorfrac{A imes i}{B} floor ]

只要考虑计算 (sum_{i=1}^{n}lfloorfrac{A imes i}{B} floor) 即可
类欧几里德算法
若 (A>B)，那么就是

[lfloorfrac{A}{B} floorsum_{i=1}^{n}i+sum_{i=1}^{n}lfloorfrac{(A mod B) imes i}{B} floor ]

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} imes i,iin[1,n]) 与坐标轴围成的三角形中的整点的个数
设 (m=lfloorfrac{A imes n}{B} floor)
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 imes gcd(A,B)}{B})，矩形内的就是 (n imes 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} imes i,iin[1,lfloorfrac{A imes n}{B} floor]) 与坐标轴围成的三角形中的整点的个数
即

[sum_{i=1}^{lfloorfrac{A imes n}{B} floor}lfloorfrac{B imes i}{A} floor ]

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 imes i-B imes sum_{i=1}^{n}lfloorfrac{A imes i}{B} floor ]

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 (lfloorfrac{n}{B} floor)，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
", Query(1, 1, cnt, l, r));
	}
	return 0;
}