AtCoder Beginner Contest 156

D - Bouquet

组合数学。
二项式定理。
注意取模之前一定要保证他是正数（有时候只加一次mod可能不够）

#include <bits/stdc++.h>
#define ll long long

using namespace std;
const int mod = 1e9 + 7;

ll qmi(ll a, ll k, ll p){
    ll res = 1;
    while(k){
        if(k & 1)
            res = (ll)res * a % p;
        a = (ll)a * a % p;
        k >>= 1;
    }
    return res;
}

ll C (ll a, ll b) {
    ll fz = 1, fm = 1;
    for (ll i = 1; i <= b; i++)    fm = fm % mod * i % mod;
    for (ll i = 0; i < b; i++)     fz = fz % mod * (a - i) % mod;
    return fz % mod * qmi (fm, mod - 2, mod) % mod;
}

int main () {
    ll n, a, b, sum;
    cin >> n >> a >> b;
    sum = qmi (2, n, mod) - 1 + mod - C(n, a) - C(n, b) + mod;
    cout << max(0ll, (sum + mod) % mod);
}

//2^n-1-C(n,a)-C(n,b)

E - Roaming

组合数学。
隔板法。
学习这种转化思想

#include <bits/stdc++.h>
#define ll long long

using namespace std;
const int N = 2e5 + 5, mod = 1e9 + 7;
ll fact[N], infact[N];

ll qmi(ll a, ll k, ll p){
    ll res = 1;
    while(k){
        if(k & 1)
            res = (ll)res * a % p;
        a = (ll)a * a % p;
        k >>= 1;
    }
    return res;
}

ll C (int a, int b) {
    if (a < b)  return 0;
    return fact[a] * infact[b] % mod * infact[a - b] % mod;
}

void init () {
    fact[0] = infact[0] = 1;
    for(int i = 1; i < N; i++){
        fact[i] = (ll)fact[i-1] * i % mod;
        infact[i] = (ll)infact[i-1] * qmi(i, mod - 2,mod) % mod;
    }
}

int main () {
    init ();
    ll n, k, sum = 0;
    cin >> n >> k;
    for (int i = 0; i <= min (k, n - 1); i++)     (sum += C(n, i) * C (n - 1, n - i - 1) % mod) %= mod;
    cout << sum;
}

//枚举空房子个数i然后隔板法(n个人分到n-i个房子中)

F - Modularness

求相反。
等于：模数相等的 \(d_i\) 的个数
大于：m进制下进位的个数

#include <bits/stdc++.h>
#define ll long long

using namespace std;
const int N = 5005;
int k, q, n, x, m;
ll a[N], d[N], d0[N]; //注意ll

int main () {
    cin >> k >> q;
    for (int i = 1; i <= k; i++)    cin >> d0[i];
    while (q--) {
        cin >> n >> x >> m;
        n --;
        ll sum = n, cnt = 0;
        for (int i = 1; i <= k; i++) {
            d[i] = d0[i] % m;
            a[i] = a[i-1] + d[i];
            cnt += (d[i] == 0); //相等的个数
        }
        sum -= n / k * cnt;
        for (int i = 1; i <= n % k; i++)    sum -= (d[i] == 0);
        //计算进位的个数
        ll r = x + n / k * a[k] + a[n % k];
        sum -= r / m - x / m;
        cout << sum << endl;
    }
}