[HAOI2012]高速公路

嘟嘟嘟


涉及到概率的题，估计我就只会这种吧……


这一类概率问题比较好想，就是每一种情况/总方案数。而这两者一般都可以分别维护。
对于这道题，因为是一条链，于是把边看成点，那么总方案数就是\(C_{R - L + 1} ^ 2\)，询问的时候直接搞出来就行。
关键是分子。很显然得先推推式子。有一个技巧就是式子中的下标尽量统一，也就是全局编号，这样就不用像我一样区间合并的时候还得考虑左右区间，然后分别维护从区间左右端点开始编号的一对乱七八糟的东西了。
考虑每一条边权的贡献，那么分子就是$$ans = \sum _ {i = l} ^ {r} a_i * (i - l + 1) * (r - i + 1)$$然后我们把式子拆开$$ans = (-l * r - l + r + 1) * \sum a_i + (l + r) \sum a_i * i + \sum a_i * i ^ 2$$到这就很简单了，发现我们只用维护这三个\(\sum\)。而这三个\(\sum\)都是极其好维护的。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
#include<assert.h>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 1e5 + 5;
In ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
In void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
In void MYFILE()
{
#ifndef mrclr
  freopen(".in", "r", stdin);
  freopen(".out", "w", stdout);
#endif
}

char s[2];
int n, m;
ll S[maxn];

In ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}

struct Tree
{
  int l, r, len;
  ll lzy, sum1, sum2, sum3;
  In Tree operator + (const Tree& oth)const
  {
    Tree ret;
    ret.l = l, ret.r = oth.r;
    ret.len = ret.r - ret.l + 1;
    ret.lzy = 0;
    ret.sum1 = sum1 + oth.sum1;
    ret.sum2 = sum2 + oth.sum2;
    ret.sum3 = sum3 + oth.sum3;
    return ret;
  }
}t[maxn << 2];
In void build(int L, int R, int now)
{
  t[now].l = L, t[now].r = R;
  t[now].len = R - L + 1;
  t[now].lzy = t[now].sum1 = t[now].sum2 = t[now].sum3 = 0;
  if(L == R) return;
  int mid = (L + R) >> 1;
  build(L, mid, now << 1);
  build(mid + 1, R, now << 1 | 1);
}
In void change(int now, ll d)
{
  t[now].lzy += d;
  t[now].sum1 += d * t[now].len;
  t[now].sum2 += d * t[now].len * (t[now].l + t[now].r) / 2;
  t[now].sum3 += d * (S[t[now].r] - S[t[now].l - 1]);
}
In void pushdown(int now)
{
  if(t[now].lzy)
    {
      change(now << 1, t[now].lzy), change(now << 1 | 1, t[now].lzy);
      t[now].lzy = 0;
    }
}
In void update(int L, int R, int now, ll d)
{
  if(L > R) return;
  if(t[now].l == L && t[now].r == R) {change(now, d); return;}
  pushdown(now);
  int mid = (t[now].l + t[now].r) >> 1;
  if(R <= mid) update(L, R, now << 1, d);
  else if(L > mid) update(L, R, now << 1 | 1, d);
  else update(L, mid, now << 1, d), update(mid + 1, R, now << 1 | 1, d);
  t[now] = t[now << 1] + t[now << 1 | 1];
}
In Tree query(int L, int R, int now)
{
  if(t[now].l == L && t[now].r == R) return t[now];
  pushdown(now);
  int mid = (t[now].l + t[now].r) >> 1;
  if(R <= mid) return query(L, R, now << 1);
  else if(L > mid) return query(L, R, now << 1 | 1);
  else return query(L, mid, now << 1) + query(mid + 1, R, now << 1 | 1);
}

int main()
{
  MYFILE();
  n = read(), m = read();
  for(int i = 1; i <= n; ++i) S[i] = S[i - 1] + 1LL * i * i;
  build(1, n - 1, 1);
  for(int i = 1; i <= m; ++i)
    {
      scanf("%s", s); int L = read(), R = read() - 1;
      if(s[0] == 'C')
	{
	  int v = read();
	  update(L, R, 1, v);
	}
      else
	{
	  if(L > R) {puts("0"); continue;}
	  int len = R - L + 1;
	  Tree tp = query(L, R, 1);
	  ll ans1 = tp.sum1 * (R - L - 1LL * L * R + 1) + tp.sum2 * (L + R) - tp.sum3;
	  ll ans2 = (1LL * (len + 1) * len) >> 1;
	  ll Gcd = gcd(ans1, ans2);
	  write(ans1 / Gcd), putchar('/'), write(ans2 / Gcd), enter;
	}
    }
  return 0;
}