赛时 rank 19,T1 0,T2 25 T3 10 T4 100
T3 挂了10pts?
数学专场，套路专场，烧脑专场。
幸亏我还有缓存的李超树博客，最后一个小时就溜了去打数据结构。
数学好难，拜谢数学。
T1 黑暗型高松灯
Company Acquisitions
要用势能分析，鞅的停时定理。由于赛时这个放T1非常逆天，所以整场比赛的奖励也十分逆天。
不会做，跳了。
T2 速度型高松灯
[HNOI2011] 数学作业
按位考虑，矩阵快速幂。
点此查看代码
#include<bits/stdc++.h>
#include<bits/extc++.h>
// using namespace __gnu_pbds;
// using namespace __gnu_cxx;
using namespace std;
#define infile(x) freopen(x,"r",stdin)
#define outfile(x) freopen(x,"w",stdout)
#define errfile(x) freopen(x,"w",stderr)
using ll=long long;using ull=unsigned long long;
using db = double;using ldb = long double;
#ifdef LOCALFILE *InFile = infile("in.in"),*OutFile = outfile("out.out");// FILE *ErrFile=errfile("err.err");
#elseFILE *Infile = stdin,*OutFile = stdout;//FILE *ErrFile = stderr;
#endif
#define int long long
int n,mod;
struct matrix{int s[3][3];matrix operator*(matrix a){matrix ans;memset(ans.s,0,sizeof ans.s);for(int i=0;i<3;i++)for(int j=0;j<3;j++)for(int k=0;k<3;k++)ans.s[i][j]=((ans.s[i][j]+a.s[i][k]*s[k][j])%mod+mod)%mod;return ans;}
}ans,base,k;
inline matrix new_one(){matrix a;memset(a.s,0,sizeof a.s);for(int i=0;i<3;i++)a.s[i][i]=1;return a;
}
inline matrix ksm(matrix a,int b){matrix ans = new_one();for(;b;b >>= 1,a = a*a)if(b&1)ans = ans*a;return ans;
}
__int128_t ten[22];
signed main(){cin.tie(nullptr)->sync_with_stdio(false);cout.tie(nullptr)->sync_with_stdio(false);cin>>n>>mod;n++;ten[0]=1;for(int i=1;i<=20;i++)ten[i]=ten[i-1]*10;memset(base.s,0,sizeof base.s);k.s[0][0]=0,k.s[1][0]=0,k.s[2][0]=1;base.s[0][2]=1,base.s[1][2]=1,base.s[2][1]=-1,base.s[2][2]=2;for(int i=1;i<=(int)ceil(1+log(n)/log(10.0));i++){base.s[0][0]=ten[i]%mod;if(n<ten[i]){k=k*ksm(base,n-ten[i-1]);return cout<<k.s[0][0],0;}k=k*ksm(base,ten[i]-ten[i-1]);}
}
T3 力量型高松灯
简单题
莫反，有个恶心的\((i+j)^k\)
\[\begin{aligned}
&\sum_{i=1}^n\sum_{j=1}^n(i+j)^k\mu^2(\gcd(i,j))\gcd(i,j)\\
=&\sum_{d=1}^n\mu^2(d)d\sum_{i=1}^n\sum_{j=1}^n(i+j)^k[\gcd(i,j)=d]\\
=&\sum_{d=1}^n\mu^2(d)d^{k+1}\sum_{i=1}^{\lfloor\frac nd\rfloor}\sum_{j=1}^{\lfloor\frac nd\rfloor}(i+j)^k[\gcd(i,j)=1]\\
=&\sum_{d=1}^n\mu^2(d)d^{k+1}\sum_{i=1}^{\lfloor\frac nd\rfloor}\sum_{j=1}^{\lfloor\frac nd\rfloor}(i+j)^k\sum_{e|i,e|j}\mu(e)\\
=&\sum_{e=1}^n\mu(e)e^k\sum_{d=1}^{\lfloor\frac ne\rfloor}\mu^2(d)d^{k+1}\sum_{i=1}^{\lfloor\frac n{de}\rfloor}\sum_{j=1}^{\lfloor\frac n{de}\rfloor}(i+j)^k\\
=&\sum_{T=1}^nS\left(\left\lfloor\frac nT\right\rfloor\right)T^k\sum_{d|T}\mu^2(d)\mu\left(\frac Td\right)d
\end{aligned}
\]
其中 \(S(n)=\sum_{i=1}^n\sum_{j=1}^n(i+j)^k\)。
考虑如何求\(S(n)\)
设\(F(n) = \sum_{i=1}^ni^k,G(n)=\sum_{i=1}^nF(i)\)，那么\(S(n)=G(2n)-2G(n)\)
线筛加前缀和就可以了
考虑后面的\(\sum_{d|T}\mu^2(d)\mu\left(\frac Td\right)d\)怎么求
我们设T有一个质因数p，若\(T=p^{3\,or\,more }\times sth.\)，那么后面的就是0。
然后就没了
点此查看代码
#include<bits/stdc++.h>
#include<bits/extc++.h>
// using namespace __gnu_pbds;
// using namespace __gnu_cxx;
using namespace std;
#define infile(x) freopen(x,"r",stdin)
#define outfile(x) freopen(x,"w",stdout)
#define errfile(x) freopen(x,"w",stderr)
using ll=long long;using ull=unsigned long long;
using db = double;using ldb = long double;
#ifdef LOCALFILE *InFile = infile("in.in"),*OutFile = outfile("out.out");// FILE *ErrFile=errfile("err.err");
#elseFILE *Infile = stdin,*OutFile = stdout;//FILE *ErrFile = stderr;
#endif
#define int long long
const int N = 1e7 + 10,mod = 998244353;
bitset<N> pd;
vector<int> prime;
int f[N],F[N],n,k;
inline int power(int a,int b,int mod){int res = 1;while(b){if(b&1) res = 1ll * res * a % mod;b >>= 1;a = 1ll * a * a % mod;}return res;
}
inline void Sieve(int n) {f[1] = F[1] = 1;for(int i = 2; i <= n; ++i) {if(!pd[i]) prime.push_back(i),f[i] = i - 1,F[i] = power(i,k,mod);for(auto j : prime) {if(i * j > n) break;int p = i * j;pd[p] = true;F[p] = 1ll * F[i] * F[j] % mod;if(i % j == 0) {int q = i / j;if(q % j) f[p] = 1ll * (mod - j) * f[q] % mod;break;} f[p] = 1ll * f[i] * (j - 1) % mod;}}for(int i = 2; i <= n; ++i) f[i] = (f[i - 1] + 1ll * f[i] * F[i] % mod) % mod,F[i] = (F[i] + F[i - 1]) % mod;for(int i = 2; i <= n; ++i) F[i] = (F[i] + F[i - 1]) % mod;
}
inline int solve(int n){return (F[n<<1] - 2ll*F[n]%mod + mod)%mod;}
signed main(){cin.tie(nullptr)->sync_with_stdio(false);cout.tie(nullptr)->sync_with_stdio(false);cin>>n>>k;k %= (mod-1);Sieve(n<<1);ll ans = 0;for(int l = 1,r;l <= n;l = r + 1){r = n / (n / l);ans = (ans + 1ll * (f[r]-f[l-1]+mod)%mod*solve(n/l)%mod)%mod;}cout<<ans;
}
T4 高松灯
贪心。
从低到高填9，计算答案即可。
特判第一位，如果后面的超限，那么就填第一位的数字-1，反之就填第一位的数字。
点此查看代码
#include<bits/stdc++.h>
#include<bits/extc++.h>
// using namespace __gnu_pbds;
// using namespace __gnu_cxx;
using namespace std;
#define infile(x) freopen(x,"r",stdin)
#define outfile(x) freopen(x,"w",stdout)
#define errfile(x) freopen(x,"w",stderr)
using ll=long long;using ull=unsigned long long;
using db = double;using ldb = long double;
#ifdef LOCALFILE *InFile = infile("in.in"),*OutFile = outfile("out.out");// FILE *ErrFile=errfile("err.err");
#elseFILE *Infile = stdin,*OutFile = stdout;//FILE *ErrFile = stderr;
#endif
signed main(){cin.tie(nullptr)->sync_with_stdio(false);cout.tie(nullptr)->sync_with_stdio(false);ll n;cin>>n;ll ans = 0;bool flag = true;while(n > 9){int x = n % 10;if(x != 9) flag = false;ans += 9;n /= 10;}ans += (flag?n:n-1);cout<<ans;
}
总结
逆天模拟赛
