[HDU] 2020中国大学生程序设计竞赛(CCPC) - 网络选拔赛 Graph Theory Class
作者:互联网
暴力很容易写,找一下规律,可以发现:
实际上不用找规律,显然\(2 \to p\),每个合数肯定是被某个质因子连上去
对于质数\(p(>3)\),首先有\(2 \to p\)的边,之后有\(p \to p \cdot (p+2k)\)的边,其中\(k=0,1,2,3,\cdots\)
最后连上\(2 \to *\),其中\(*\) 从未被连过
那就很容易发现,\(*\)只能是偶数或者素数
现在需要求mst
如果\(p+k=a \cdot b,a<p\),那么\(p(p+k)=a \cdot pb=a \cdot (a+2k)\)
所以每个奇数\(t\)只会在其最小质因子处统计,这些边的贡献是\(t\)
\(2 \to p\)的贡献是\(2p\)
\[\left( \sum_{2<p\le n+1,p \in \mathbb{P}} 2p \right)+\sum_{2 \le t \le n+1, t \not\in \mathbb{P}} t=\left( \sum_{i=3}^{n+1}i \right)+\sum_{2 < p \le n+1 ,p \in \mathbb{P}} p \]懒得写min25筛了,网上贺了个板子
#include <bits/stdc++.h>
using namespace std;
int calc(int n) {
vector<int> fa(n + 1);
function<int(int)> getfa = [&] (int x) {
return x == fa[x] ? x : fa[x] = getfa(fa[x]);
};
for(int i = 1 ; i <= n ; ++ i) {
fa[i] = i;
}
int res = 0;
function<int(int, int)> gcd = [&] (int a, int b) {
return b ? gcd(b, a % b) : a;
};
auto lcm = [&] (int a, int b) {
return a / gcd(a, b) * b;
};
vector<pair<int, pair<int, int> > > ed;
for(int i = 1 ; i <= n ; ++ i) {
for(int j = i + 1 ; j <= n ; ++ j) {
ed.push_back({ lcm(i + 1, j + 1), { i, j } });
}
}
vector<pair<pair<int, int>, int> > out;
sort(ed.begin(), ed.end());
for(auto e: ed) {
int w = e.first;
int u = e.second.first, v = e.second.second;
if(getfa(u) != getfa(v)) {
// printf("%d - %d (%d)\n", v + 1, u + 1, w);
// printf("%d %d %d\n", v + 1, u + 1, w);
out.push_back({ {u + 1, v + 1}, w });
fa[getfa(u)] = getfa(v);
res += w;
}
}
// sort(out.begin(), out.end());
// for(auto x: out) {
// printf("%d %d %d\n", x.first.first, x.first.second, x.second);
// }
return res;
}
int calc_test(int n) {
vector<int> vis(n + 5), fa(n + 5), linked(n + 5);
function<int(int)> getfa = [&] (int x) {
return x == fa[x] ? x : fa[x] = getfa(fa[x]);
};
for(int i = 2 ; i <= n + 1 ; ++ i) {
fa[i] = i;
}
function<int(int, int)> gcd = [&] (int a, int b) {
return b ? gcd(b, a % b) : a;
};
auto lcm = [&] (int a, int b) {
return a / gcd(a, b) * b;
};
int res = 0;
for(int i = 2 ; i <= n + 1 ; i += 2) vis[i] = 1;
for(int i = 3 ; i <= n + 1 ; ++ i) {
if(!vis[i]) {
linked[i] = 1;
res += lcm(2, i);
for(int j = i ; j <= n + 1 ; j += i) vis[j] = 1;
for(int j = i ; i * j <= n + 1 && j <= n + 1 ; j += 2) {
int t = i * j;
if(!linked[t]) {
linked[t] = 1;
// cout << i << ' ' << t << endl;
res += lcm(i, t);
}
}
}
}
for(int i = 3 ; i <= n + 1 ; ++ i) {
if(!linked[i]) {
// cout << 2 << ' ' << i << endl;
res += lcm(2, i);
}
}
return res;
}
int calc_test_test(int n) {
vector<int> vis(n + 5);
int res = 0;
for(int i = 2 ; i <= n + 1 ; ++ i) {
if(i > 2) {
res += i;
}
if(!vis[i]) {
if(i != 2) {
res += i;
}
for(int j = i ; j <= n + 1 ; j += i) {
vis[j] = 1;
}
}
}
return res;
}
typedef long long ll;
ll mod;
namespace getP {
// http://www.shallowdreams.cn/?p=266
// 不想写了,粘的板子
const int N = 1000010;
int prime[N], id1[N], id2[N], flag[N], ncnt, m;
ll g[N], sum[N], a[N], T;
ll n;
int ID(ll x) {
return x <= T ? id1[x] : id2[n / x];
}
ll calc(ll x) {
return x * (x + 1) / 2 - 1;
}
ll f(ll x) {
return x;
}
ll init(ll n){
T = sqrt(n + 0.5);
for (int i = 2; i <= T; i++) {
if (!flag[i]) prime[++ncnt] = i, sum[ncnt] = sum[ncnt - 1] + i;
for (int j = 1; j <= ncnt && i * prime[j] <= T; j++) {
flag[i * prime[j]] = 1;
if (i % prime[j] == 0) break;
}
}
for (ll l = 1; l <= n; l = n / (n / l) + 1) {
a[++m] = n / l;
if (a[m] <= T) id1[a[m]] = m; else id2[n / a[m]] = m;
g[m] = calc(a[m]);
}
for (int i = 1; i <= ncnt; i++)
for (int j = 1; j <= m && (ll)prime[i] * prime[i] <= a[j]; j++)
g[j] = g[j] - (ll)prime[i] * (g[ID(a[j] / prime[i])] - sum[i - 1]);
}
ll solve(ll x){
if(x<=1){return x;}
memset(g,0,sizeof(g));
memset(a,0,sizeof(a));
memset(sum,0,sizeof(sum));
memset(prime,0,sizeof(prime));
memset(id1,0,sizeof(id1));
memset(id2,0,sizeof(id2));
memset(flag,0,sizeof(flag));
ncnt=m=0;
return n=x,init(n),g[ID(n)];
}
}
typedef long long ll;
ll nekkonya(ll n) {
// 2 ~ n + 1
if(n == 1) return 0;
else if(n == 2) return 6;
else if(n == 3) return 10;
else if(n == 4) return 20;
else if(n == 5) return 26;
ll res = getP :: solve(n + 1) - 2;
res %= mod;
res += (ll) (3 + n + 1) % mod * ((n - 1) % mod) % mod * ((mod + 1) / 2) % mod;
res %= mod;
res = (res + mod) % mod;
return res;
}
int main() {
int t; cin >> t;
while(t --) {
ll n;
cin >> n >> mod;
cout << nekkonya(n) << endl;
}
}
标签:HDU,return,gcd,int,Graph,CCPC,fa,res,getfa 来源: https://www.cnblogs.com/nekko/p/15388085.html