D、数列求和(嘤雄难度) J、滑稽树下你和我 I、滑稽树上滑稽果

D、数列求和(嘤雄难度)

1、 $A_i=i*i+i$

2、 $S_A_i=\frac{n*(n+1)*(2*n+1)}{6}+ \frac {n*(n+1)}{2}=\frac{n*(n+1)*(n+2)}{3}$

3、考虑先求出 $m$ 的所有质因数，然后通过容斥来求所有与 $m$ 不互质的 $A_i$ 的和。

4、假设当前容斥算的质数是 $k$ ，那么就有 $\frac{n}{k}$ 个该质数的倍数，即要求

$\sum _{i=1} ^{\frac{n}{k}} {ki*ki+ki}$ ，式子化简得 $k*k*\sum _{i=1} ^{\frac{n}{k}} {i*i}+k*\sum _{i=1} ^{\frac{n}{k}} {i}$ ，求和公式再化简即可在 $O(1)$ 时间内算出结果。

Code：

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const ll mod = 1e9 + 7;
vector<ll>v;
ll power(ll a, ll b)
{ll res = 1;while (b){if (b & 1)res = res * a % mod;a = a * a % mod;b >>= 1;}return res;
}
void get(ll k)
{for (int i = 2; i * i <= k; i++){if (k % i == 0){v.push_back(i);while (k % i == 0)k /= i;}}if (k > 1)v.push_back(k);
}
ll n, m;
ll inv2 = power(2, mod - 2);
ll inv6 = power(6, mod - 2);
ll calc(ll k)
{ll cnt = n / k;ll res = ((k * k % mod) * (cnt * (cnt + 1) % mod) % mod) * ((2 * cnt + 1) * inv6 % mod) % mod + (cnt * (cnt + 1) % mod) * (k * inv2 % mod) % mod;return res % mod;
}
int main()
{while (scanf("%lld%lld", &n, &m) > 0){v.clear();get(m);int sz = v.size();ll ans = 0;for (int i = 0; i < (1 << sz); i++){int num = 1, cishu = 0;for (int j = 0; j < sz; j++){if (i & (1 << j)){num *= v[j];cishu++;}}if (cishu & 1)ans = (ans - calc(num) + mod) % mod;elseans = (ans + calc(num)) % mod;}printf("%lld\n", ans);}
}

J、滑稽树下你和我

计算每条边的贡献，即两边点的个数的乘积，然后再乘这条边的长度，总和就是答案

Code:

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int MAXN = 1e5 + 5;
const ll mod = 1e9 + 7;
struct edge
{int to;ll w;int nex;
}e[MAXN * 2];
int head[MAXN], tot = 1;
void add(int a, int b, ll c)
{e[tot] = edge{ b,c,head[a] };head[a] = tot++;
}
int num[MAXN], n;
bool vis[MAXN];
void init()
{memset(head, -1, sizeof(head));tot = 1;
}
ll ans = 0;
#define Pair pair<int,int>
map<Pair,ll>mp;
void dfs(int u, int fa)
{num[u] = 1;for (int i = head[u]; i + 1; i = e[i].nex){int v = e[i].to;if (v != fa){dfs(v, u);num[u] += num[v];ll ad = 1LL * (n - num[v]) * num[v] % mod;if (u < v)ad *= mp[Pair{ u,v }];elsead *= mp[Pair{ v,u }];ad %= mod;ans = (ans + ad) % mod;}}
}
int main()
{init();int a, b;ll c;scanf("%d", &n);for (int i = 1; i < n; i++){scanf("%d%d%lld", &a, &b, &c);add(a, b, c);add(b, a, c);if (a > b)swap(a, b);mp[Pair{ a,b }] = c;}dfs(1, 0);printf("%lld\n", ans);
}

I、滑稽树上滑稽果

莫队算法，考虑将所求的作为单独一项即 $\sum _{i=0} ^{m} {n \choose i}$ ，为了方便，下面将使用 $S_{C_n^m}$ 来表示 $\sum_{i=0}^m {n \choose i}$ 。

预处理阶乘和阶乘的逆元

Code:

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const ll mod = 1e9 + 7;
const int MAXN = 1e5 + 5;
struct node
{int index;int n;int m;
}in[MAXN];
int id[105];
ll fac[MAXN], inv[MAXN];
ll ans[MAXN];
ll power(ll a, ll b)
{ll res = 1;while (b){if (b & 1)res = res * a % mod;a = a * a % mod;b >>= 1;}return res;
}
void init()
{fac[0] = fac[1] = 1;for (int i = 2; i < MAXN; i++)fac[i] = fac[i - 1] * i % mod;inv[0] = inv[1] = 1;inv[100000] = power(fac[100000], mod - 2);for (int i = 99999; i > 1; i--)inv[i] = inv[i + 1] * (i + 1) % mod;
}
ll comb(ll a, ll b)
{if (a < b || b < 0)return 0;ll res = fac[a] * inv[b] % mod * inv[a - b] % mod;return res;
}
ll inv2 = (mod + 1) / 2;
int unit;
int main()
{init();int num;scanf("%d", &num);for (int i = 0; i < num; i++){in[i].index = i;scanf("%d%d", &in[i].n, &in[i].m);}unit = sqrt(num);sort(in, in + num, [](node q, node w) {if (q.n / unit == w.n / unit)return q.m < w.m;return q.n < w.n;});int a = 0, b = -1;ll res = 0;for (int i = 0; i < num; i++){while (a < in[i].n){res = (res * 2 - comb(a, b) + mod) % mod;a++;}while (a > in[i].n){res = ((res + comb(a - 1, b)) % mod * inv2) % mod;a--;}while (b < in[i].m){res = (res + comb(a, b + 1)) % mod;b++;}while (b > in[i].m){res = (res - comb(a, b) + mod) % mod;b--;}ans[in[i].index] = res * power(inv2, in[i].n) % mod;}for (int i = 0; i < num; i++)printf("%lld\n", ans[i]);
}