数论总结

1 void exgcd(ll a,ll b,ll &d,ll &x,ll &y)
2 {
3     if(!b) x=1,y=0,d=a;
4     else {exgcd(b,a%b,d,x,y);int t=x;x=y;y=t-a/b*y;}
5 }

 1 for(int i=2;i<=n-1;i++)
 2 {
 3         if(!b[i])
 4         {
 5             prime[++cnt]=i;
 6             phi[i]=i-1;
 7         }
 8         for(int j=1;j<=cnt&&i*prime[j]<=n-1;j++)
 9         {
10             b[i*prime[j]]=1;
11             if(i%prime[j]==0)
12             {
13                 phi[i*prime[j]]=phi[i]*prime[j];
14                 break;
15             }
16             phi[i*prime[j]]=phi[i]*(prime[j]-1);
17         }
18 }

 1 ll qpow(ll b)
 2 {   
 3     ll a=10,ret=1;
 4     while(b)
 5     {
 6        if(b%2==1) ret=ret*a%n;
 7        a=a*a%n; 
 8        b=b>>1;
 9     }
10     return ret;
11 }

 1 #include<algorithm>
 2 #include<iostream>
 3 #include<cstring>  
 4 #include<vector>  
 5 using namespace std;  
 6 typedef long long ll;  
 7 const int MOD=10000;  
 8 typedef vector<ll> vec;  
 9 typedef vector<vec> matrix;  
10 matrix mul(matrix &a,matrix &b)
11 {  
12     matrix c(a.size(),vec(b[0].size()));  
13     for(int i=0;i<2;++i)  
14     {  
15         for(int j=0;j<2;++j)  
16         {  
17             for(int k=0;k<2;++k)  
18             {  
19                 c[i][j]+=a[i][k]*b[k][j];  
20                 c[i][j]%=MOD;  
21             }  
22         }  
23     }  
24     return c;  
25 }  
26 matrix qpow(matrix a,ll n)  
27 {  
28     matrix res(a.size(),vec(a.size()));  
29     for(int i=0;i<a.size();++i)  
30     res[i][i]=1; 
31     while(n)  
32     {  
33         if(n%2==1) res=mul(res,a);  
34         a=mul(a,a);  
35         n=n>>1;  
36     }  
37     return res;  
38 }  
39 ll solve(ll n)  
40 {  
41     matrix a(2,vec(2));  
42     a[0][0]=1;  
43     a[0][1]=1;  
44     a[1][0]=1;  
45     a[1][1]=0;  
46     a=qpow(a,n);  
47     return a[0][1];
48 }  
49 int main()  
50 {  
51     ll n;
52     cin>>n;
53     while(n!=-1)
54     {
55         cout<<solve(n)<<endl;
56         cin>>n;
57     }
58     return 0;  
59 }

 1 #include<iostream>
 2 #include<algorithm>
 3 using namespace std;
 4 const int N=100005;
 5 typedef long long ll;
 6 ll t,n,m,p,a[N];
 7 inline ll qpow(ll y,ll z,ll p)
 8 {
 9     y%=p;
10     ll ans=1;
11     for(int i=z;i;i>>=1,y=y*y%p)if(i&1)ans=ans*y%p;
12     return ans;
13 }
14 inline ll C(ll n,ll m)
15 {
16     if(m>n)return 0;
17     return ((a[n]*qpow(a[m],p-2,p))%p*qpow(a[n-m],p-2,p)%p);
18 }
19 inline ll Lucas(int m,int n)
20 {
21     if(n==0)
22     return 1;
23     return Lucas(m/p,n/p)%p*C(m%p,n%p);
24 }
25 int main()
26 {
27     p=5000011;
28     cin>>m>>n;
29     a[0]=1;
30     for(int i=1;i<=p;++i)
31     a[i]=(a[i-1]*i)%p;
32     cout<<Lucas(m+n,n)%p<<endl;
33     return 0;
34 }

 1 int main()
 2 {
 3     int n, cnt = 0;
 4     int prime[100001];
 5     bool vis[100001];
 6     scanf("%d", &n);
 7     memset(vis, false, sizeof(vis));
 8     memset(prime, 0, sizeof(prime));
 9     for(int i = 2; i <= n; i++)
10     {
11         if(!vis[i])
12         prime[cnt++] = i;
13         for(int j = 0; j<cnt && i*prime[j]<=n; j++)
14         {
15             vis[i*prime[j]] = true;
16             if(i % prime[j] == 0) break;
17         }
18     }
19     printf("%d\n", cnt);
20 }

1 int BashGame(int n,int m)
2 {
3     if(n%(m+1)) return 0;
4     return 1;
5 }

1 int NimGame(int n)
2 {
3     int flag=0;
4     for(int i=0;i<n;i++)
5     flag^=f[i];
6     if(flag)return 1;
7     return 0;
8 }

1 int WGame（int a,int b)
2 {
3     a=min(a,b);
4     b=max(a,b);
5     int k=b-a;
6     double path=(sqrt(5)+1)/2;
7     if(a==floor(path*k) return 1;
8     return 0;
9 }

 1 int FibonacciGame(int n)
 2 {
 3     if(n==1) return 1;
 4     int f[1e5];
 5     f[0]=f[1]=1;
 6     for(int i=2;f[i-1]<n;i++)
 7     {
 8         f[i]=f[i-1]+f[i-2];
 9         if(f[i]==n) return 1;
10     }
11     return 0;
12 }