Codeforces Round #258 (Div. 2)Devu and Flowers 容斥原理

#include <bits\stdc++.h>

using namespace std;

const int MOD = 1e9 + 7;
typedef long long LL;

LL invv[25];
LL inv(LL x)/// 求逆元
{
    return x == 1 ? 1LL : (MOD - MOD / x) * inv (MOD % x) % MOD;
}
LL Cmn(LL n, LL m) ///求组合数
{
    LL ret = 1;
    for (int i = 1; i <= m; i++)
        ret = (n - i + 1) % MOD * ret % MOD * invv[i] % MOD;
    return ret;
}
int calc(int x)
{
    int ret = 0;
    while (x)
    {
//        x -= x & (-x);
        x=x&(x-1);
        ret ^= 1;
    }
    return ret;
}
int n;
LL s, a[25];
int main()
{
    for (int i = 1; i < 25; i++) invv[i] = inv(i);
    cin >> n >> s;
    for (int i = 0; i < n; i++)
        cin >> a[i];
    int ALL = (1 << n) - 1;
    LL ans = 0;
    for (int i = 0; i <= ALL; i++)///遍历所有集合
    {
        LL nows = s;
        int num = 0;///统计当前集合元素个数，以确定符号
        for (int j = 0; j < n; j++)
        {
            if (i & (1 << j))
            {
                nows -= a[j] + 1;
                num++;
            }
        }
        if (nows < 0) continue;
        if (num & 1)///集合含有奇数个元素，符号为-
            ans -= Cmn(nows + n - 1, n - 1);
        else///集合含有偶数个元素，符号为+
            ans += Cmn(nows + n - 1, n - 1);
        ans %= MOD;
    }
    cout << (ans + MOD) % MOD << endl;
    return 0;
}