51Nod 1667 概率好题 - 容斥原理

题目传送门

　　无障碍通道

　　有障碍通道

题目大意

　　若$L_{i}\leqslant x_{i} \leqslant R_{i}$，求$\sum x_{i} = 0$以及$\sum x_{i} < 0$的方案数。$(L_{i}R_{i} \geqslant 0)$（好吧。是概率）

　　听完题解感觉自己是个傻逼。组合数学白学了。

　　如果$L_{i} \neq 0$，那么取$a_{i} = x_{i} - L_{i}$。

　　然后容斥。

　　如何处理$\sum x_{i} < 0$？加一个物品$0\leqslant x_{n + 1} < \infty $。

　　然后做完了。

Code

 /**
  * 51nod
  * Problem#1667
  * Accepted
  * Time: 187ms
  * Memory: 2052k
  */
 #include <bits/stdc++.h>
 using namespace std;
 typedef bool boolean;

 const int M = 1e9 + ;
 const signed int inf = (signed) (~0u >> );

 void exgcd(int a, int b, int& x, int& y) {
     if (!b)
         x = , y = ;
     else {
         exgcd(b, a % b, y, x);
         y -= (a / b) * x;
     }
 }

 int inv(int a, int n) {
     int x, y;
     exgcd(a, n, x, y);
     return (x < ) ? (x + n) : (x);
 }

 int add(int a, int b) {
     a += b;
     if (a < )
         a += M;
     if (a >= M)
         a -= M;
     return a;
 }

 int invs[];
 int n, m;
 int K = ;
 int rw = , req = , rl, rall = ;
 int ar[];

 inline void prepare() {
     for (int i = ; i <= ; i++)
         invs[i] = inv(i, M);
 }

 inline void init() {
     scanf("%d", &n);
     rw = req = , rall = , K = ;
     for (int i = , l, r; i < n; i++)
         scanf("%d%d", &l, &r), ar[i] = r - l, K -= l, rall = rall * 1ll * (ar[i] + ) % M;
     scanf("%d", &m);
     for (int i = , l, r; i < m; i++)
         scanf("%d%d", &l, &r), ar[n + i] = r - l, K += r, rall = rall * 1ll * (ar[n + i] + ) % M;
     n += m;
 }

 int C(int n, int m) {
     int rt = ;
     for (int i = ; i < m; i++)
         rt = rt * 1ll * (n - i) % M;
     for (int i = ; i <= m; i++)
         rt = rt * 1ll * invs[i] % M;
     return rt;
 }

 int calc(int dep, int sign, int K) {
     if (K < )
         return ;
     if (dep == -)
         return sign * C(K + n - , n - );
     return add(calc(dep - , -sign, K - ar[dep] - ), calc(dep - , sign, K));
 }

 inline void solve() {
     req = calc(n - , , K);
     ar[n++] = inf;
     rw = calc(n - , , K);
     rl = add(rall, -rw);
     rw = add(rw, -req);
 //    cerr << rl << " " << req << " " << rw << endl;
     rw = rw * 1ll * inv(rall, M) % M;
     req = req * 1ll * inv(rall, M) % M;
     rl = rl * 1ll * inv(rall, M) % M;
     printf("%d %d %d\n", rl, req, rw);
 }

 int T;
 int main() {
     scanf("%d", &T);
     prepare();
     while (T--) {
         init();
         solve();
     }
     return ;
 }