B. Basic Gcd Problem

传送门

打表发现 F(n, c) = c ^ (n可以分解为多少个质因数)

用欧拉筛打出可以分解为多少个质因数

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll inf = 0x3f3f3f3f3f3f3f3f;
const double PI = acos(-1.0);
const ll mod = 1e9 + 7;
const int N = 1e6 + 10;

int pri[N], tot, cnt[N];
bool vis[N];
void init() {
    memset(vis, 0, sizeof(vis));
    memset(cnt, 0, sizeof(cnt));
    vis[0] = vis[1] = 1;
    tot = 0;
    for(int i = 2; i < N; ++i) {
        if(!vis[i]) {
            pri[++tot] = i;
            cnt[i]++;
        }
        for(int j = 1; j <= tot; ++j) {
            if(i * pri[j] > N) break;
            vis[i * pri[j]] = 1;
            cnt[i * pri[j]] = cnt[i] + 1;
        }
    }
}

ll qpow(ll a, ll b) {
    ll ans = 1;
    while(b) {
        if(b & 1)
            ans = ans * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return ans % mod;
}

int main() {
    init();
    int t, n, c;
    scanf("%d", &t);
    while(t--) {
        scanf("%d%d", &n, &c);
        cout<<qpow(1ll * c, 1ll * cnt[n]) % mod<<'\n';
    }
    return 0;
}

F. Finding the Order

传送门

题意：已知直线AB与直线CD平行，给出AC、AD、BC、BD，判断C、D的位置关系

思路1：四边形对角线之和大于两条对边和，比较AC + BD和AD + BC，大的是对角线

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll inf = 0x3f3f3f3f3f3f3f3f;
const double PI = acos(-1.0);
const ll mod = 1e9 + 7;
const int N = 1e6 + 10;

int main() {
    int t, ac, ad, bc, bd;
    scanf("%d", &t);
    while(t--) {
        scanf("%d%d%d%d", &ac, &ad, &bc, &bd);
        if(ac + bd < ad + bc)
            cout<<"AB//CD"<<'\n';
        else
            cout<<"AB//DC"<<'\n';
    }
    return 0;
}

 思路2：先判断C、D和AB中垂线的关系，AC < BC时C在AB中垂线左，AC > BC时C在中垂线右，D点同理；若两点都在左侧，比较BC、BD；若两点都在右侧，比较AC、AD

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll inf = 0x3f3f3f3f3f3f3f3f;
const double PI = acos(-1.0);
const ll mod = 1e9 + 7;
const int N = 1e6 + 10;

int main() {
    int t, ac, ad, bc, bd;
    scanf("%d", &t);
    while(t--) {
        scanf("%d%d%d%d", &ac, &ad, &bc, &bd);
        int c = 0, d = 0;   //左1 中0 右-1
        if(ac < bc) c = 1;
        else if(ac > bc) c = -1;
        if(ad < bd) d = 1;
        else if(ad > bd) d = -1;
        if(c > d)
            cout<<"AB//CD"<<'\n';
        else if(c < d)
            cout<<"AB//DC"<<'\n';
        else {
            if(c == 1) {
                if(bc > bd)
                    cout<<"AB//CD"<<'\n';
                else
                    cout<<"AB//DC"<<'\n';
            }
            else {
                if(ac < ad)
                    cout<<"AB//CD"<<'\n';
                else
                    cout<<"AB//DC"<<'\n';
            }
        }
    }
    return 0;
}

H. Harder Gcd Problem

传送门

题意：数1~n最多能组成多少对不互质的数对

思路：逆序遍历小于 n 的所有素数，取出1~n中所有该素数的倍数，两两配对，如果是奇数个留下数 p * 2，因为这个数可以最后遍历2的倍数时配。比赛的时候没想到这一点，用优先队列维护的有效（还没有被遍历过）素因子最多的数，把这个数留下来，比较麻烦，但也不是很慢：

emm还是用正解做吧：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll inf = 0x3f3f3f3f3f3f3f3f;
const double PI = acos(-1.0);
const ll mod = 1e9 + 7;
const int N = 2e5 + 10;

int pri[N], tot;
bool vis[N], is[N];
void init() {
    memset(vis, 0, sizeof(vis));
    vis[0] = vis[1] = 1;
    tot = 0;
    for(int i = 2; i < N; ++i) {
        if(!vis[i]) {
            pri[++tot] = i;
            for(int j = i + i; j < N; j += i)
                vis[j] = 1;
        }
    }
}

int main() {
    init();
    int t, n;
    scanf("%d", &t);
    while(t--) {
        scanf("%d", &n);
        int it = upper_bound(pri + 1, pri + tot + 1, n) - pri- 1;
        vector<int>vec;
        memset(is, 0, sizeof(is));
        for(int i = it; i >= 1; --i) {
            int cnt = 1;
            is[pri[i]] = 1;
            vec.push_back(pri[i]);
            for(int j = pri[i] * 3; j <= n; j += pri[i]) {
                if(!is[j]) {
                    is[j] = 1;
                    cnt++;
                    vec.push_back(j);
                }
            }
            if(cnt & 1) {
                if(pri[i] * 2 <= n && !is[pri[i] * 2]) {
                    vec.push_back(pri[i] * 2);
                    is[pri[i] * 2] = 1;
                }
                else
                    vec.pop_back();
            }
        }
        int ans = vec.size() / 2;
        cout<<ans<<'\n';
        for(int i = 0; i < ans * 2; i += 2)
            cout<<vec[i]<<' '<<vec[i + 1]<<'\n';
        vec.clear();
    }
    return 0;
}