线段树+概率
今天这道题爆零了,奥妙重重。
其实我们可以把式子化成这样:sigma((i-l+1)*(r-i+1)*ai) 这里r减了1
然后展开,(1-l)*(r+1)*ai+(r+l)*i*ai-i*i*ai
我们发现除了含有i的项其他都可以提到外面,也就是说我们要维护ai,i*ai,i*i*ai三个量。
那么就用线段树维护,打标记时用数列公式即可。
#include<bits/stdc++.h> using namespace std; typedef long long ll; const int N = 100010; int n, m; struct seg { ll tag[N << 2], tree[N << 2][4]; void pushdown(int x, int l, int r) { if(!tag[x]) return; tag[x << 1] += tag[x]; tag[x << 1 | 1] += tag[x]; int mid = (l + r) >> 1; tree[x << 1][1] += tag[x] * (ll)(mid - l + 1ll); tree[x << 1 | 1][1] += tag[x] * (ll)(r - mid); tree[x << 1][2] += tag[x] * (ll)(l + mid) * (ll)(mid - l + 1) / 2ll; tree[x << 1 | 1][2] += tag[x] * (ll)(r + mid + 1ll) * (ll)(r - mid) / 2ll; tree[x << 1][3] += tag[x] * ((ll)mid * (ll)(mid + 1ll) * (ll)(2ll * mid + 1ll) - (ll)(l - 1ll) * (ll)l * (ll)(2ll * l - 1ll)) / 6ll; tree[x << 1 | 1][3] += tag[x] * ((ll)r * (ll)(r + 1ll) * (ll)(2ll * r + 1ll) - (ll)mid * (ll)(mid + 1ll) * (ll)(2ll * mid + 1ll)) / 6ll; tag[x]= 0ll; } void update(int l, int r, int x, int a, int b, ll delta) { if(l > b || r < a) return; if(l >= a && r <= b) { tag[x] += delta; tree[x][1] += (ll)(r - l + 1ll) * delta; tree[x][2] += (ll)(l + r) * (ll)(r - l + 1) / 2ll * delta; tree[x][3] += ((ll)r * (ll)(r + 1ll) * (2ll * r + 1) - (ll)(l - 1ll) * (ll)l * (ll)(2ll * l - 1ll)) / 6ll * delta; return; } pushdown(x, l, r); int mid = (l + r) >> 1; update(l, mid, x << 1, a, b, delta); update(mid + 1, r , x << 1 | 1, a, b, delta); for(int i = 1; i <= 3; ++i) tree[x][i] = tree[x << 1][i] + tree[x << 1 | 1][i]; } ll query(int l, int r, int x, int a, int b, int type) { if(l > b || r < a) return 0ll; if(l >= a && r <= b) return tree[x][type]; pushdown(x, l, r); int mid = (l + r) >> 1; return (query(l, mid, x << 1, a, b, type) + (query(mid + 1, r, x << 1 | 1, a, b, type))); } } t; ll gcd(ll a, ll b) { return !b ? a : gcd(b, a % b); } int main() { // freopen("c.in", "r", stdin); // freopen("c.out", "w", stdout); scanf("%d%d", &n, &m); --n; while(m--) { char s[10]; int l, r; ll delta; scanf("%s", s); if(s[0] == ‘C‘) { scanf("%d%d%lld", &l , &r, &delta); --r; t.update(1, n, 1, l, r, delta); } if(s[0] == ‘Q‘) { scanf("%d%d", &l, &r); --r; ll T1 = t.query(1, n, 1, l, r, 1); ll T2 = t.query(1, n, 1, l, r, 2); ll T3 = t.query(1, n, 1, l, r, 3); ll ans = (ll)(1ll - l) * (ll)(r + 1ll) * T1 + (ll)(r + l) * T2 - T3; ll T4 = (ll)(r - l + 2ll) * (ll)(r - l + 1ll) / 2ll; ll T = gcd(ans, T4); printf("%lld/%lld\n", ans / T, T4 / T); } } // fclose(stdin); fclose(stdout); return 0; }
时间: 2024-11-03 08:30:02