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打表出c的存在没什么作用。
容易找出规律 答案就是c的某个次数，这个次数就是n的质因子的幂次和。
因此只需线性筛出幂次和，再快速幂就好了
#pragma warning(disable:4996) #include<iostream> #include<algorithm> #include<bitset> #include<tuple> #include<unordered_map> #include<fstream> #include<iomanip> #include<string> #include<cmath> #include<cstring> #include<vector> #include<map> #include<set> #include<list> #include<queue> #include<stack> #include<sstream> #include<cstdio> #include<ctime> #include<cstdlib> #define INF 0x3f3f3f3f #define inf 0x7FFFFFFF #define MOD 1000000007 #define moD 1000000003 #define pii pair<ll,int> #define eps 1e-7 #define equals(a,b) (fabs(a-b)<eps) #define bug puts("bug") #define re register #define fi first #define se second #define pb push_back const int maxn = 1e6 + 5; const double Inf = 10000.0; const double PI = acos(-1.0); typedef long long ll; typedef unsigned long long ull; using namespace std; ll num[maxn]; int vis[maxn]; int num_sieve(int n) { int cnt = 0; for (int i = 2; i <= n; i++) { if (!vis[i]) num[i] = 1, vis[i] = 1; int j = 2; while (j * i <= maxn) if (vis[j]) num[i * j] = num[i] + num[j],vis[j * i] = 1,j++; else break; } return cnt; //返回x小于等于n的s素数的个数 } ll quickPower(ll a, ll b, ll m) { ll ans = 1; ll base = a; while (b) { if (b & 1) { ans *= base; ans %= m; } base *= base; base %= m; b >>= 1; } return ans; } int main() { int T; num_sieve(maxn - 3); ll n, c; scanf("%d", &T); while (T--) { scanf("%lld%lld", &n, &c); if (n == 1ll) { puts("1"); continue; } ll cnt = 0; // cout << cnt << endl; printf("%lld ", quickPower(c, num[n] , MOD)); } }
H 。
给出n，要求从n种分出两组a，b，使ab两两满足gcd > 1
1.显然1和对于p * 2 > n 的p不可能是答案。
2.对每个数取最大质因子成为一个集合。
若集合个数是偶数，可以直接匹配。
否则，把2p扔进另一组数，剩余的可以两两匹配。
#pragma warning(disable:4996) #include<iostream> #include<algorithm> #include<bitset> #include<tuple> #include<unordered_map> #include<fstream> #include<iomanip> #include<string> #include<cmath> #include<cstring> #include<vector> #include<map> #include<set> #include<list> #include<queue> #include<stack> #include<sstream> #include<cstdio> #include<ctime> #include<cstdlib> #define INF 0x3f3f3f3f #define inf 0x7FFFFFFF #define MOD 1000000007 #define moD 1000000003 #define pii pair<int,int> #define eps 1e-7 #define equals(a,b) (fabs(a-b)<eps) #define bug puts("bug") #define re register #define fi first #define se second #define pb push_back const int maxn = 5e5 + 5; const double Inf = 10000.0; const double PI = acos(-1.0); typedef long long ll; typedef unsigned long long ull; using namespace std; int prime[maxn]; int is_prime[maxn]; int cnt; int pfac[maxn]; void is_Prime() { is_prime[0] = is_prime[1] = 1; for (int i = 2; i <= maxn -3; i++) { if (!is_prime[i]) { prime[cnt++] = i; pfac[i] = i; for (int j = 2 * i; j <= maxn - 3; j += i) { is_prime[j] = 1; pfac[j] = i; } } } } vector<int> p[maxn]; int main() { int T; int m; is_Prime(); scanf("%d", &T); while (T--) { memset(p, 0, sizeof p); scanf("%d", &m); vector<pii> ans; vector<int> tmp; for (int i = 2; i <= m; i++) { if (!is_prime[i] && i * 2 > m) continue; p[pfac[i]].push_back(i); } for (int i = 0;; i++) { if (prime[i] > m) break; if (p[prime[i]].empty()) continue; if (p[prime[i]].size() == 1) continue; if (p[prime[i]].size() & 1) { tmp.push_back(p[prime[i]][1]); ans.emplace_back(p[prime[i]][0], p[prime[i]][2]); for (int k = 3; k + 1< p[prime[i]].size(); k += 2) { ans.emplace_back(p[prime[i]][k], p[prime[i]][k + 1]); } } else { for (int k = 0; k + 1 < p[prime[i]].size(); k += 2) ans.emplace_back(p[prime[i]][k], p[prime[i]][k + 1]); } } for (int i = 0; i + 1 < tmp.size(); i+=2) { ans.emplace_back(tmp[i], tmp[i + 1]); } printf("%d ", ans.size()); for (auto it : ans) { printf("%d %d ", it.fi, it.se); } } }