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  1. #include <bits/stdc++.h>
  2. using namespace std;
  3.  
  4. typedef long long ll;
  5.  
  6. const int N = 1111;
  7. const ll M = 1e9 + 9999;
  8.  
  9. int n;
  10. ll L[N+5], R[N+5], fact[N+5], ifact[N+5];
  11.  
  12. inline ll add(ll a, ll b) { a += b; return (a >= M) ? a - M : a; }
  13. inline ll sub(ll a, ll b) { a -= b; return (a < 0) ? a + M : a; }
  14. inline ll mul(ll a, ll b) { return (__int128)a * b % M; }
  15.  
  16. ll pw(ll x, ll n) {
  17.     ll a = 1;
  18.     x %= M;
  19.     for (; n; n >>= 1, x = mul(x, x))
  20.         if (n & 1) a = mul(a, x);
  21.     return a;
  22. }
  23.  
  24. ll inv(ll x) { return pw(x, M - 2); }
  25.  
  26. void precompute(int n) {
  27.     fact[0] = 1;
  28.     for (int i = 1; i <= n; i++)
  29.         fact[i] = mul(fact[i - 1], i);
  30.     ifact[n] = inv(fact[n]);
  31.     for (int i = n; i > 0; i--)
  32.         ifact[i - 1] = mul(ifact[i], i);
  33. }
  34.  
  35. ll lagrange(const vector<ll>& y, ll n) {
  36.     int k = (int)y.size() - 1;
  37.     if (n <= k) return y[n];
  38.  
  39.     static ll p[N+5], s[N+5];
  40.     p[0] = 1;
  41.     for (int i = 0; i <= k; i++)
  42.         p[i + 1] = mul(p[i], sub(n, i));
  43.     s[k + 1] = 1;
  44.     for (int i = k; i >= 0; i--)
  45.         s[i] = mul(s[i + 1], sub(n, i));
  46.  
  47.     ll ans = 0;
  48.     for (int i = 0; i <= k; i++) {
  49.         ll num = mul(p[i], s[i + 1]), den = mul(ifact[i], ifact[k - i]);
  50.         ll term = mul(y[i], mul(num, den));
  51.         if ((k - i) & 1) ans = sub(ans, term);
  52.         else ans = add(ans, term);
  53.     }
  54.     return ans;
  55. }
  56.  
  57. int main(){
  58.     ios_base::sync_with_stdio(0); cin.tie(0);
  59.  
  60.     cin >> n;
  61.     precompute(n + 1);
  62.     for (int i = 0; i < n; i++)
  63.         cin >> L[i] >> R[i];
  64.  
  65.     vector<ll> pts;
  66.     for (int i = 0; i < n; i++) {
  67.         pts.push_back(L[i]);
  68.         pts.push_back(R[i] + 1);
  69.     }
  70.     sort(pts.begin(), pts.end());
  71.     pts.erase(unique(pts.begin(), pts.end()), pts.end());
  72.  
  73.     ll ans = 0;
  74.     for (int id = 0; id < (int)pts.size() - 1; id++) {
  75.         ll l = pts[id], r = pts[id + 1] - 1;    
  76.         if (l > r) continue;
  77.  
  78.         vector<int> active;
  79.         ll base = 1;
  80.         bool skip = false;
  81.  
  82.         for (int i = 0; i < n; i++)
  83.             if (R[i] < l) base = mul(base, R[i] - L[i] + 1);
  84.             else if (L[i] > r) {
  85.                 skip = true;
  86.                 break;
  87.             } else active.push_back(i);
  88.  
  89.         if (skip) continue;
  90.  
  91.         int deg = active.size();
  92.         int K = deg + 1;
  93.         vector<ll> T(K + 1), A(K + 1);
  94.         for (int x = 0; x <= K; x++) {
  95.             ll val = base;
  96.             for (int idx : active)
  97.                 val = mul(val, x + l - L[idx]);
  98.             A[x] = val;
  99.         }
  100.  
  101.         T[0] = 0;
  102.         for (int x = 1; x <= K; x++) {
  103.             ll w = mul(x + l - 1, sub(A[x], A[x - 1]));
  104.             T[x] = add(T[x - 1], w);
  105.         }
  106.  
  107.         ans = add(ans, lagrange(T, r - l + 1));
  108.     }
  109.  
  110.     cout << ans << "\n";
  111.  
  112.     cerr << "TIME: " << 1.0 * clock() / CLOCKS_PER_SEC << " s\n";
  113.     return 0;
  114. }
  115.  
Success #stdin #stdout #stderr 0.01s 5316KB
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TIME: 0.005904 s
https://ideone.com/QEzSzv
language:
C++14 (gcc 8.3)
created:
6 hours ago
visibility:
public

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