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
*/