Codeforces Round #258 (Div. 2)Devu and Flowers Inclusion and Exclusion Principle_html/css_WEB-ITnose

Title: Codeforces Round #258 (Div. 2) Devu and Flowers Question meaning: n boxes, the i-th box has fi flowers, the flowers in each box are exactly the same, and the flowers in different boxes are different, find the following The number of options for removing s flowers from n boxes. n<=20, s<=1e14, fi<=1e12. For permutation and combination questions, the inclusion exclusion principle can be used to solve the problem. There are 2 ways to write dfs and collection. The following is how to write a collection.

#include 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;}