20191101

A

题意

求一个三元组\((a,b,c)\)，使得 \(a * b * c \% p = 1\)。

做法

暴力记录有每一个模值多少个数，枚举即可。

#include<bits/stdc++.h>
using namespace std;
#pragma GCC optimize(3, "Ofast")
#define int long long
map <int,int> p;
map <int,int> g;
map <int,int> t;
int n;
int f[3000]; int pl;
int P;
int inv;
int getinv(int x){
	int xt = x;
	int ret = 1;
	int base = P - 2;
	while (base >= 1){
		if (base & 1) ret = xt * ret % P;
		xt = xt * xt % P;
		base /= 2;
	}
	return ret;
}
signed main(){
	freopen("awesome.in","r",stdin);
	freopen("awesome.out","w",stdout);

	cin >> n >> P;
	int ans = 0;
	for (int i=1;i<=n;i++){
		int tp;
		cin >> tp;
		if (tp % P == 0) continue;
		p[tp] ++;
//		if (p[tp] == 3){
//			if (tp * tp % P * tp % P == 1) ans ++; // deal with 3
//		}
	}
	typedef map<int,int>::iterator Iter;
	for (Iter it = p.begin(); it != p.end(); it ++){
			f[++pl] = it -> first;
	}
	for (int i=1;i<=pl;i++){
		t[f[i] % P] ++;
	}
	for (int j=pl;j>=1;j--){
		for (int i=1;i<j;i++){
			int inv = getinv(f[i] * f[j] % P);
			if (g.count(inv)) ans += g[inv];
		}
		if (p[f[j]] != 1){
			int inv = getinv(f[j] * f[j] % P);
			if (t.count(inv)) ans += t[inv];//, cout << f[j] << ' ' << f[j] << endl;
			if (p[f[j]] == 2 && f[j] * f[j] % P * f[j] % P == 1) ans --;//, cout << '?' << endl;
		}
		g[f[j] % P] ++;
	} // deal with 1 & 2 & 3
	cout << ans << endl;
}

B

最长上升子序列Ex

二分时改一下边界即可

关键代码：

        int n = read();
        for (int i = 1; i <= n; ++i)
            its[i].wei = read(), its[i].val = read();
        sort(its + 1, its + n + 1);
        int m = read();
        for (int i = 1; i <= m; ++i)
            lim[i] = read();
        sort(lim + 1, lim + m + 1, Comp);
        for (int i = 1; i <= n + 2; ++i)
            _max[i] = 0;
        int ans = 0;
        for (int i = 1; i <= n && ans <= m; ++i){
            int l = 1, r = i, mid, bans = 0;
            while (l <= r){
                mid = l + r >> 1;
                if (_max[mid] >= its[i].val) l = mid + 1, bans = mid;
                else r = mid - 1;
            }
            f[i] = bans + 1;
            if (its[i].wei > lim[f[i]])
                f[i] = -1000000007;
            if (f[i] > ans) ans = f[i];
            if (f[i] > 0){
                if (_max[f[i]] < its[i].val)
                    _max[f[i]] = its[i].val;
            }
        }

C

一道动归题。

令 \(f_{i,j}\) 表示当子树大小为 \(i\) 时深度小于等于 \(j\) 的数量。

对于一个新的点， \(f_{i,j} = \sum_k f_{k,j-1} C_{i-2}^{k-1} * f_{i-k,j}\)

边界： \(f_{1,i} = 1\)

关键代码：

    for (int i = 1; i <= n; i++) 
		f[1][i] = 1;
    for (int i = 2; i <= n; i++)
        for (int j = 1; j <= n; j++)
            for (int k = 1; k <= i - 1; k++)
                if (!ban[k])
                    f[i][j] = (f[i][j] + 1ll * f[k][j - 1] * C(i - 2, k - 1) % Mod * f[i - k][j]) % Mod;