AtCoder Beginning Contest 156

不会组合数学被吊打了……

题目链接：https://atcoder.jp/contests/abc156

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A：

/* basic header */
#include <bits/stdc++.h>
/* define */
#define ll long long
#define pb emplace_back
#define mp make_pair
#define eps 1e-8
#define lson (curpos<<1)
#define rson (curpos<<1|1)
/* namespace */
using namespace std;
/* header end */

int n, r;

int main() {
    cin >> n >> r;
    if (n >= 10) cout << r << endl;
    else cout << r + 100 * (10 - n) << endl;
    return 0;
}

B：

/* basic header */
#include <bits/stdc++.h>
/* define */
#define ll long long
#define pb emplace_back
#define mp make_pair
#define eps 1e-8
#define lson (curpos<<1)
#define rson (curpos<<1|1)
/* namespace */
using namespace std;
/* header end */

int n, k;

int main() {
    cin >> n >> k;
    int ans = 0;
    while (n) {
        ans++;
        n /= k;
    }
    cout << ans << endl;
    return 0;
}

C：

注意是任意位置，所以要枚举到100

/* basic header */
#include <bits/stdc++.h>
/* define */
#define ll long long
#define pb emplace_back
#define mp make_pair
#define eps 1e-8
#define lson (curpos<<1)
#define rson (curpos<<1|1)
/* namespace */
using namespace std;
/* header end */

const int maxn = 110;
int n, a[maxn], ans = INT_MAX;

int main() {
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i];
    for (int i = 1; i <= 100; i++) {
        int tmp = 0;
        for (int j = 1; j <= n; j++) tmp += (a[j] - i) * (a[j] - i);
        ans = min(ans, tmp);
    }
    cout << ans << endl;
    return 0;
}

D：

题意是有n朵互不相同的花，从中选择至少1朵组成一束花，且花的朵数不等于a或b，有多少种选法。

显然C(n,1)+C(n,2)+...+C(n,n)=2^n-1, 答案就是2^n-1-C(n,a)-C(n,b)

/* basic header */
#include <bits/stdc++.h>
/* define */
#define ll long long
#define pb emplace_back
#define mp make_pair
#define eps 1e-8
#define lson (curpos<<1)
#define rson (curpos<<1|1)
/* namespace */
using namespace std;
/* header end */

const ll mod = 1e9 + 7;
ll n, a, b;

ll qp(ll x, ll b) {
    ll ret = 1, base = x;
    while (b) {
        if (b & 1) ret = ret * base % mod;
        base = base * base % mod;
        b >>= 1;
    }
    return ret;
}

ll C(ll n, ll m) {
    if (!m) return 1LL;
    if (m == 1) return n;
    ll ret = 1, maxx = max(m, (n - m)), minn = min(m, (n - m));
    for (ll i = n; i > maxx; i--) ret = ret * i % mod;
    for (ll i = minn; i >= 1; i--) {
        ll currInv = qp(i, mod - 2);
        ret = ret * currInv % mod;
    }
    return ret;
}

int main() {
    cin >> n >> a >> b;
    ll total = (qp(2, n) % mod - 1 % mod + mod) % mod;
    ll c1 = C(n, a) % mod, c2 = C(n, b) % mod;
    ll ans = ((total - c1 + mod) % mod - c2 + mod) % mod;
    cout << ans << endl;
    return 0;
}

E：

题意是有n间房子，一开始每间房子里各有一个人，定义一次移动：某人从房间i走到房间j，且i不等于j。现在已知房间个数n和已经发生的移动次数k。问：房间人数的可能情况有多少种？

ans=∑ C(n, i) * C(n - 1, n - i - 1), 1<=i<=k

i的意思是选定i间空房子。由于有n个人，这n个人会形成n-1个空隙，要在这n-1个空隙里插入n-i-1个隔板来形成n-i个非空房子。

C(n - 1, n - i - 1)可以O(n)预处理出来。

/* basic header */
#include <bits/stdc++.h>
/* define */
#define ll long long
#define pb emplace_back
#define mp make_pair
#define eps 1e-8
#define lson (curpos<<1)
#define rson (curpos<<1|1)
/* namespace */
using namespace std;
/* header end */

const int maxn = 2e5 + 10;
const ll mod = 1e9 + 7;
ll fac[maxn + 1], inv[maxn + 1], dp[maxn + 1], n, k;

ll qp(ll x, ll y = mod - 2) {
    ll ret = 1;
    for (; y; y >>= 1, x = x * x % mod)
        if (y & 1) ret = ret * x % mod;
    return ret;
}

ll C(int x, int y) {
    return fac[x] * inv[y] % mod * inv[x - y] % mod;
}

int main() {
    fac[0] = fac[1] = 1;
    for (int i = 1; i <= maxn; i++) fac[i] = fac[i - 1] * i % mod;
    inv[maxn] = qp(fac[maxn]);
    for (int i = maxn - 1; i >= 0; i--) inv[i] = inv[i + 1] * (i + 1) % mod;
    cin >> n >> k;
    k = min(n - 1, k);
    ll ans = 1;
    for (int i = 1; i <= n; i++) dp[i] = C(n - 1, i - 1);
    for (int i = 1; i <= k; i++) ans = (C(n, i) * dp[n - i] % mod + ans) % mod;
    cout << ans << endl;
    return 0;
}