51Nod 1667 概率好题

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$L_{i} eq 0$，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$0leqslant 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 + 7;
const signed int inf = (signed) (~0u >> 1);

void exgcd(int a, int b, int& x, int& y) {
    if (!b)
        x = 1, y = 0;
    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 < 0) ? (x + n) : (x);
}

int add(int a, int b) {
    a += b;
    if (a < 0)
        a += M;
    if (a >= M)
        a -= M;
    return a;
}

int invs[25];
int n, m;
int K = 0;
int rw = 0, req = 0, rl, rall = 0;
int ar[20];

inline void prepare() {
    for (int i = 1; i <= 20; i++)
        invs[i] = inv(i, M);
}

inline void init() {
    scanf("%d", &n);
    rw = req = 0, rall = 1, K = 0;
    for (int i = 0, l, r; i < n; i++)
        scanf("%d%d", &l, &r), ar[i] = r - l, K -= l, rall = rall * 1ll * (ar[i] + 1) % M;
    scanf("%d", &m);
    for (int i = 0, l, r; i < m; i++)
        scanf("%d%d", &l, &r), ar[n + i] = r - l, K += r, rall = rall * 1ll * (ar[n + i] + 1) % M;
    n += m;
}

int C(int n, int m) {
    int rt = 1;
    for (int i = 0; i < m; i++)
        rt = rt * 1ll * (n - i) % M;
    for (int i = 1; i <= m; i++)
        rt = rt * 1ll * invs[i] % M;
    return rt;
}

int calc(int dep, int sign, int K) {
    if (K < 0)
        return 0;
    if (dep == -1)
        return sign * C(K + n - 1, n - 1);
    return add(calc(dep - 1, -sign, K - ar[dep] - 1), calc(dep - 1, sign, K));
}

inline void solve() {
    req = calc(n - 1, 1, K);
    ar[n++] = inf;
    rw = calc(n - 1, 1, 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
", rl, req, rw);
}

int T;
int main() {
    scanf("%d", &T);
    prepare();
    while (T--) {
        init();
        solve();
    }
    return 0;
}