题目来源：Light OJ 1411 Rip Van Winkle`s Code

题意：3中操作 1种查询 求区间和 其中每次可以把一段区间从左到右加上1，2，3，。。。或者从右到左加上。。。3，2，1 或者把某个区间的数都置为v

思路：我是加了6个域 

add是这段区间每个数都要加上add  add是这么来的 对与123456。。。这个等差数列 可能要分为2个区间 那么我就分成123和123 两个右边的等差数列每个数还应该加上3 所以右区间add加3

v是这个区间都要置为v 他的优先级最高 

b是代表这个区间有多少个递增的等差数列

c是代表这个区间有多少个递减的等差数列

sum是区间和 

f为真说明要执行C 因为v可以是正和负还可以是0

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 250010;
typedef long long LL;
struct node
{
	LL sum, add, v, b, c;
 	bool f;
}a[maxn<<2];

void build(int l, int r, int rt)
{
	a[rt].sum = 0;
	a[rt].add = 0;
	a[rt].b = 0;
	a[rt].c = 0;
	a[rt].f = false;
	if(l == r)
		return;
	int m = (l + r) >> 1;
	build(l, m, rt<<1);
	build(m+1, r, rt<<1|1);
}

void pushdown(int rt, int l, int r)
{
	LL k = (LL)(r-l+1);
	//printf("%d %d %lld\n", l, r, a[rt].v);
	if(a[rt].f)
	{
		a[rt<<1].sum = (LL)a[rt].v*(k-(k>>1));
		a[rt<<1|1].sum = (LL)a[rt].v*(k>>1);
		a[rt<<1].v = a[rt<<1|1].v = a[rt].v;
		a[rt<<1].f = a[rt<<1|1].f = true;
		a[rt].f = false;
		a[rt<<1].b = a[rt<<1|1].b = 0;
		a[rt<<1].c = a[rt<<1|1].c = 0;
		a[rt<<1].add = a[rt<<1|1].add = 0;
		
	}
	LL s1 = (k-(k>>1));
	LL s2 = (k>>1);
	if(a[rt].b)
	{
		a[rt<<1].sum += (LL)a[rt].b*(1+s1)*s1/2;
		a[rt<<1|1].sum += (LL)a[rt].b*(1+s2)*s2/2+(LL)s2*s1*a[rt].b;
		a[rt<<1].b += a[rt].b;
		a[rt<<1|1].b += a[rt].b;
		a[rt<<1|1].add += (LL)s1*a[rt].b;	
		a[rt].b = 0;
	}
	if(a[rt].c)
	{
		a[rt<<1].sum += (LL)a[rt].c*(1+s1)*s1/2+(LL)s2*s1*a[rt].c;
		a[rt<<1|1].sum += (LL)a[rt].c*(1+s2)*s2/2;
		a[rt<<1].c += a[rt].c;
		a[rt<<1|1].c += a[rt].c;
		a[rt<<1].add += (LL)s2*a[rt].c;
		a[rt].c = 0;
	}
	if(a[rt].add)
	{
		a[rt<<1].sum += (LL)a[rt].add*(k-(k>>1));
		a[rt<<1|1].sum += (LL)a[rt].add*(k>>1);
		a[rt<<1].add += a[rt].add;
		a[rt<<1|1].add += a[rt].add; 
		a[rt].add = 0;
	}
}
void update(int x, int y, int l, int r, int rt, LL add, LL v, char c)
{
	//puts("we");
	if(l == x && r == y)
	{	
		if(c == 'A')
		{
			/*if(a[rt].f)
			{
				a[rt].sum = (LL)(r-l+1)*v;
				//a[rt].f = false;
			}*/
			a[rt].sum += (LL)(r-l+1)*(r-l+1+1)/2;
			a[rt].b++;
			a[rt].sum += (LL)add*(r-l+1);
			a[rt].add += add;
			//printf("%d %d %d %lld\n", l, r, a[rt].b, a[rt].sum);
		}
		else if(c == 'B')
		{
			/*if(a[rt].f)
			{
				a[rt].sum = (LL)(r-l+1)*v;
				//a[rt].f = false;
			}*/
			a[rt].sum += (LL)(r-l+1)*(r-l+1+1)/2;
			a[rt].c++;
			a[rt].sum += (LL)add*(r-l+1);
			a[rt].add += add;
		}
		else if(c == 'C')
		{
			a[rt].f = true;
			a[rt].add = 0;
			a[rt].sum = (LL)(r-l+1)*v;
			a[rt].b = 0;
			a[rt].c = 0;
			a[rt].v = v;
			
		}
		return;
	}
	int m = (l + r) >> 1;
	pushdown(rt, l, r);
	if(y <= m)
	{
		update(x, y, l, m, rt<<1, add, v, c);
	}
	else if(x > m)
	{
		update(x, y, m+1, r, rt<<1|1, add, v, c);
	}
	else
	{
		if(c == 'A')
		{
			update(x, m, l, m, rt<<1, add, v, c);
			update(m+1, y, m+1, r, rt<<1|1, add+(m-x+1), v, c);
		}
		else if(c == 'B')
		{
			update(x, m, l, m, rt<<1, add+(y-m), v, c);
			update(m+1, y, m+1, r, rt<<1|1, add, v, c);
		}
		else if(c == 'C')
		{
			//puts("as");
			update(x, m, l, m, rt<<1, add, v, c);
			update(m+1, y, m+1, r, rt<<1|1, add, v, c);
		}
	}
	a[rt].sum = a[rt<<1].sum + a[rt<<1|1].sum;
	
}

LL query(int x, int y, int l, int r, int rt)
{
	if(x == l && y == r)
	{
		//printf("**%d %d %lld\n", l, r, a[rt].sum);
		return a[rt].sum;
	}
	pushdown(rt, l, r);
	int m = (l + r) >> 1;
	LL ans = 0;
	if(y <= m)
		ans += query(x, y, l, m, rt<<1);
	else if(x > m)
		ans += query(x, y, m+1, r, rt<<1|1);
	else
		ans = query(x, m, l, m, rt<<1) + query(m+1, y, m+1, r, rt<<1|1);
	a[rt].sum = a[rt<<1].sum + a[rt<<1|1].sum;
	return ans;
}
int main()
{
	int cas = 1;
	int T;
	scanf("%d", &T);
	while(T--)
	{
		int n = 250000;
		int q;
		scanf("%d", &q);
		printf("Case %d:\n", cas++);
		build(1, n, 1);
		while(q--)
		{
			char s[10];
			scanf("%s", s);
			if(s[0] == 'A')
			{
				
				int x, y;
				scanf("%d %d", &x, &y);
				update(x, y, 1, n, 1, 0, 0, 'A');
			}
			else if(s[0] == 'B')
			{
				int x, y;
				scanf("%d %d", &x, &y);
				update(x, y, 1, n, 1, 0, 0, 'B');
			}
			else if(s[0] == 'C')
			{
				int x, y;
				LL w;
				scanf("%d %d %lld", &x, &y, &w);
				update(x, y, 1, n, 1, 0, w, 'C');
			}
			else
			{
				int x, y;
				scanf("%d %d", &x, &y);
				printf("%lld\n", query(x, y, 1, n, 1));
			}
		}
		
	}
	return 0;
} 
/*
10
B 1 4
A 1 4
S 2 2
B 1 3
C 1 4 1
B 1 4
C 3 3 3
A 2 3
S 2 3
B 3 3
*/