T1
(n le 20 ,m le n!)
这道题可以证明一定有解，我们每次找出 比我们要凑成的数小且是n!的约数 里最大的数就可以了。跑得飞快~
#include<iostream>
#include<cstring>
#include<cstdio>
#define LL long long
using namespace std;
LL n, m, jc, tot, js;
const LL N = 3700010, inf = 0x3f3f3f3f3f3f3f3f;
LL w[N], cnt[N], ans[N];
inline int read()
{
int res = 0; char ch = getchar(); bool XX = false;
for (; !isdigit(ch); ch = getchar())(ch == '-') && (XX = true);
for (; isdigit(ch); ch = getchar())res = (res << 3) + (res << 1) + (ch ^ 48);
return XX ? -res : res;
}
void fen()
{
LL t = jc;
for (int i = 2; i * i <= t; ++i)
if (!(t % i))
{
w[++tot] = i;
while (!(t % i))++cnt[tot], t /= i;
}
if (t != 1)w[++tot] = t, ++cnt[tot];
}
LL dfs2(LL now, LL ans, LL top)
{
if (now == tot + 1)return ans;
if (ans > top)return 0;
LL res = 0;
for (int i = 0; i <= cnt[now]; ++i)
{
res = max(res, dfs2(now + 1, ans, top));
ans *= w[now];
if (ans > top)break;
}
return res;
}
void dfs(LL x)
{
if (!(jc % x)) {ans[++js] = x; return;}
LL t = dfs2(1, 1, x);
ans[++js] = t;
dfs(x - t);
}
signed main()
{
cin >> n >> m; jc = 1;
for (int i = 1; i <= n; ++i)jc *= i;
fen(); dfs(m);
cout << js << endl;
for (int i = 1; i <= js; ++i)printf("%lld ", ans[i]);
fclose(stdin); fclose(stdout);
return 0;
}
T2
(n le 10^6)
这道题之前做过，其实就是把求有多少个不同的子串放到了树上，我们只需要在建后缀自动机的时候把id[fa[x]]当成last就行了，统计的时候ans += len[i] - len[fa[i]];就行了.
#include<iostream>
#include<cstdio>
#include<map>
#define LL long long
using namespace std;
int n, tot, x, y, cnt;
LL ans;
const int N = 1000010;
int head[N], du[N], fa[N << 1], id[N], len[N << 1];
map<int, int>tr[N << 1];
struct bian {int to, nt;} e[N << 1];
void add(int f, int t)
{
e[++tot] = (bian) {t, head[f]};
head[f] = tot;
}
inline int read()
{
int res = 0; char ch = getchar(); bool XX = false;
for (; !isdigit(ch); ch = getchar())(ch == '-') && (XX = true);
for (; isdigit(ch); ch = getchar())res = (res << 3) + (res << 1) + (ch ^ 48);
return XX ? -res : res;
}
void Insert(int x, int Q, int las)
{
int np = ++cnt, p = las, q, nq;
len[np] = len[p] + 1; id[Q] = cnt;
for (; p && tr[p].find(x) == tr[p].end(); p = fa[p])tr[p][x] = np;
if (!p)fa[np] = 1;
else if (len[q = tr[p][x]] == len[p] + 1)fa[np] = q;
else
{
len[nq = ++cnt] = len[p] + 1;
fa[nq] = fa[q]; fa[q] = fa[np] = nq; tr[nq] = tr[q];
for (; p; p = fa[p])
{
if (tr[p].find(x) == tr[p].end() || tr[p][x] != q)break;
tr[p][x] = nq;
}
}
}
void dfs(int x, int fa)
{
Insert(du[x], x, id[fa]);
for (int i = head[x]; i; i = e[i].nt)
if (e[i].to != fa)dfs(e[i].to, x);
}
int main()
{
cnt = id[0] = 1; cin >> n;
for (int i = 1; i < n; ++i)
x = read(), y = read(), add(x, y), add(y, x), ++du[x], ++du[y];
dfs(1, 0);
for (int i = 2; i <= cnt; ++i)ans += len[i] - len[fa[i]];
cout << ans;
return 0;
}
T3
提交答案题，这里就不讲了...
下午
对于任意的积性函数 (f),有(f(gcd(a,b)) imes f(lcm(a,b)) = f(a) imes f(b))。然后就变成了(displaystyle (sum_{i=1}^nf(i))^2)
然后就是杜教筛了，(g)为(id),(f*g=1),证明.
注意取模
#include<iostream>
#include<cstdio>
#include<map>
#define LL long long
using namespace std;
LL n, tot, ans;
const int mod = 1e9 + 7, inv2 = (mod + 1) >> 1, N = 1000010, M = 1000000;
int vis[N], zhi[N], f[N];
map<LL, LL>mp;
void YYCH()
{
vis[1] = f[1] = 1;
for (int i = 2; i <= M; ++i)
{
if (!vis[i])zhi[++tot] = i, f[i] = 1 - i;
for (int j = 1; j <= tot && i * zhi[j] <= M; ++j)
{
vis[i * zhi[j]] = 1;
if (!(i % zhi[j])) {f[i * zhi[j]] = f[i]; break;}
f[i * zhi[j]] = f[i] * f[zhi[j]];
}
}
for (int i = 2; i <= M; ++i)(f[i] += f[i - 1]) %= mod;
}
LL Sum(LL l, LL r) {return (l + r) * (r - l + 1) % mod * inv2 % mod;}
LL S(LL n)
{
if (n <= M)return f[n];
if (mp.count(n))return mp[n];
LL res = 0;
for (LL l = 2, r; l <= n; l = r + 1)
{
r = n / (n / l);
(res += Sum(l, r) % mod * S(n / l) % mod) %= mod;
}
return mp[n] = (n - res) % mod;
}
signed main()
{
cin >> n;
YYCH();
ans = S(n);
cout << (ans * ans % mod + mod) % mod;
fclose(stdin); fclose(stdout);
return 0;
}