Minimum spanning tree HDU - 6954

题意：

给定n-1个点，编号从2到n，两点a和b之间的边权重为lcm（a，b）。请找出它们形成的最小生成树。
2<=n<=10000000

题解：

这题一看就眼熟。。。这不是去年的CCPC网络赛吗，当时就差这个题进区域赛，CCPC里面数据范围是n<=10 ^10 , 这个是10 ^7,前者用min25筛做，后者直接用欧拉筛就可以
HDU 6889 Graph Theory Class(CCPC网络赛)

代码：

min25筛做法
min25筛模板出处

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N = 1000010;typedef long long ll;ll qpow(ll a, ll b)
{ll ans = 1;while(b){if(b & 1)ans = ans * a;a = a * a ;b /= 2;}return ans;
}ll prime[N], id1[N], id2[N], flag[N], ncnt, m;ll g[N], sum[N], a[N], T, n;inline ll ID(ll x)
{return x <= T ? id1[x] : id2[n / x];
}inline ll calc(ll x)
{return x * (x + 1) / 2 - 1;
}inline ll f(ll x)
{return x;
}inline void init()
{ncnt = m = 0;T = sqrt(n + 0.5);for (ll i = 2; i <= T; i++){if (!flag[i])prime[++ncnt] = i, sum[ncnt] = sum[ncnt - 1] + i;for (ll 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;elseid2[n / a[m]] = m;g[m] = calc(a[m]);}for (ll i = 1; i <= ncnt; i++)for (ll 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]);
}inline ll solve(ll x)
{if (x <= 1)return x;return n = x, init(), g[ID(n)];
}int main()
{ll n, t;scanf("%lld", &t);while(t--) {scanf("%lld", &n);n--;ll ans = (solve(n + 1) - 2 );//质数和 ll tmp = ((n + 4)  * (n-1) )  /2;printf("%lld\n", (ans + tmp) );}
}

欧拉筛做法：