A 【模板】快速幂
板子，略
cpp
#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll a,p,k;
int main()
{
scanf("%lld%lld%lld",&a,&p,&k);
printf("%lld^%lld mod %lld=",a,p,k);
ll ans=1,w=a;a%=k;
while(p)
{
if(p&1)
{
ans=ans*w%k;
}
w=w*w%k;
p>>=1;
}printf("%lld",ans%k);
}
cpp
#include<bits/stdc++.h>
using namespace std;
long long mod=1000000007;
long long n,k;
struct sq{
long long num[102][102];
sq(){memset(num,0,sizeof(num));}//初始化num数组为空
};
sq operator *(const sq &a,const sq &b)
{
sq ans;
for(int k=1;k<=n;k++)
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
ans.num[i][j]=(ans.num[i][j]+a.num[i][k]*b.num[k][j]%mod)%mod;
}
}return ans;
}//一个重载
sq x,y,ans;
long long read()
{
char s;long long ans=0;
s=getchar();
while(s>'9'||s<'0'){s=getchar();}
while(s>='0'&&s<='9')
{
ans=ans*10+s-'0';
s=getchar();
}
return ans;
}//只是个快速读入函数
inline void init()
{
n=read();k=read();//用scanf,cin也行
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
x.num[i][j]=read();
for(int i=1;i<=n;i++)
{
y.num[i][i]=1;
ans.num[i][i]=1;
}
}
int main(){
init();
while(k)
{
if(k&1){ans=ans*x;}
x=x*x;
k>>=1;
}
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
printf("%lld ",ans.num[i][j]);
printf("\n");
}
return 0;
}除以一个数等于乘以它的逆元，逆元可以用费马小定理求
cpp
#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll a,b,mod=19260817;
inline ll read()
{
ll ans=0;char s=getchar();
while(s>'9'||s<'0')s=getchar();
while(s>='0'&&s<='9')
{
ans=((ans<<1)%mod+(ans<<3)%mod+(s&15))%mod;
s=getchar();
}return ans;
}
ll quick_m(ll i,ll n)
{
i%=mod;
ll ans=1,ds=i;
while(n)
{
if(n&1)
{
ans=ans*ds%mod;
}
ds=ds*ds%mod;
n>>=1;
}return ans%mod;
}
signed main(){
a=read();b=read();
if(b==0){printf("Angry!");return 0;}
printf("%lld",a*quick_m(b,mod-2)%mod);
return 0;
}
cpp
#include<bits/stdc++.h>
int a;
int main()
{
scanf("%d",&a);
for(int i=2;i*i<=a;i++)
{
if(a%i==0)
{
printf("%d",std::max(a/i,i));
return 0;//找到一个因数就可以停了，另一个因数就是最大的那个
}
}
}数据不大的话，每个数都直接用质数判断法也能过
cpp
#include<bits/stdc++.h>
using namespace std;
int n,a;
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
scanf("%d",&a);
bool f=1;
for(int j=2;j*j<=a;j++)
{
if(a%j==0){f=0;break;}
}
if(f==1&&a!=1)printf("%d ",a);//是质数
}
}这里用的欧拉筛
cpp
#include<bits/stdc++.h>
using namespace std;
bool ans[100000002];
int z[6000005],p;
int n,m,a;
int main()
{
scanf("%d%d",&n,&m);
for(int i=2;i<=n;i++)
{
if(!ans[i])z[++p]=i;//printf("[%d]",z[p]);
for(int j=1;j<=p&&z[j]*i<=n;j++)
{
ans[i*z[j]]=1;
if(i%z[j]==0)break;
}
}
for(int i=1;i<=m;i++)
{
scanf("%d",&a);
printf("%d\n",z[a]);
}
return 0;
} 求三个数的最小公倍数
cpp
#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll gcd(ll x,ll y){return !y?x:gcd(y,x%y);}
ll lcm(ll x,ll y){return x*y/gcd(x,y);}
int main()
{
ll a,b,c;
scanf("%lld%lld%lld",&a,&b,&c);
printf("%lld",lcm(a,lcm(b,c)));
}H 最大公约数和最小公倍数问题
x ∗ y = g c d ( x , y ) ∗ l c m ( x , y ) x*y=gcd(x,y)*lcm(x,y) x∗y=gcd(x,y)∗lcm(x,y)
枚举x然后判断算出对应y是否满足gcd,lcm就行，可能要特判
cpp
#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll gcd(ll x,ll y){return !y?x:gcd(y,x%y);}
ll lcm(ll x,ll y){return x*y/gcd(x,y);}
int main()
{
ll a,b,ans=0;
scanf("%lld%lld",&a,&b);
if(a==b)ans--;//这样会存在x,y相同的一组，下面枚举时会多算一组，故删去
for(ll i=1;i*i<=a*b;i++)
if(a*b%i==0&&gcd(i,a*b/i)==a)ans+=2;
printf("%lld",ans);
}建议看我之前博客线性递推式部分
a − 1 = − ⌊ p / a ⌋ ∗ ( p m o d a ) − 1 ( m o d p ) a^{-1}=-⌊p/a⌋*(p \ mod \ a )^{-1} \ \ \ \ (mod \ p) a−1=−⌊p/a⌋∗(p mod a)−1 (mod p)
cpp
#include<bits/stdc++.h>
#define ll long long
using namespace std;
int n,p,inv[3000005];
signed main(){
scanf("%d%d",&n,&p);
inv[0]=1;inv[1]=1;
printf("1\n");
for(int i=2;i<=n;i++)
{
inv[i]=(ll)inv[p%i]*(p-p/i)%p;//-p/i不能直接取模，要+p变成正数再%p
printf("%d\n",inv[i]);
}
return 0;
}
/*下面是欧拉筛版
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N=3000005;
ll n,p;
bool vis[N];
ll z[N];
ll inv[N];
ll quick(ll a,ll k,ll mod)
{
ll ans=1,as=a;
while(k)
{
if(k&1)
{
ans=ans*as%mod;
}
as=as*as%mod;
k>>=1;
}
return ans;
}
int main()
{
scanf("%lld%lld",&n,&p);
vis[1]=1,inv[1]=1;
for (int i=2;i<=n;i++)
{
//printf("(%d\n",i);
if (!vis[i])z[++z[0]]=i,inv[i]=quick(i,p-2,p);
for (int j=1;j<=z[0]&&i*z[j]<=n;j++)
{
vis[i*z[j]]=1;
inv[i*z[j]]=(inv[i]*inv[z[j]])%p;
if (i%z[j]==0) break;
}
}
for (int i=1;i<=n;i++)
printf("%lld\n",inv[i]);
return 0;
}*/求 1 − > N ! 1->N! 1−>N!中与 M ! M! M!互质的个数
推演一下，最后得 N ! M ! ϕ ( M ! ) \frac {N!}{M!} \phi{(M!)} M!N!ϕ(M!)
=> N ! ( 1 − p 1 p 1 ) ( 1 − p 2 p 2 ) . . . ( 1 − p k p k ) N!(\frac {1-p_1}{p_1})(\frac {1-p_2}{p_2})...(\frac {1-p_k}{p_k}) N!(p11−p1)(p21−p2)...(pk1−pk)
一堆预处理，质数+线性求逆+阶乘。。。
cpp
#include<bits/stdc++.h>
#define ll long long
using namespace std;
inline ll read()
{
ll ans=0;char s=getchar();
while(!isdigit(s))s=getchar();
while(isdigit(s))
{
ans=(ans*10)+(s&15);
s=getchar();
}return ans;
}
const int N=1e7+10;
int T;
ll n,m,mod;
int z[N],p;
bool vis[N];
void get_prime()
{
for(int i=2;i<N;i++)
{
if(!vis[i])z[++p]=i;
for(int j=1;j<=p&&i*z[j]<N;j++)
{
vis[i*z[j]]=1;
if(i%z[j]==0)break;
}
}
}
ll inv[N],mul[N],pt[N];
int main()
{
scanf("%d%lld",&T,&mod);
get_prime();
inv[1]=1;inv[0]=1;
for(int i=2;i<N;i++)
inv[i]=inv[mod%i]*(mod-mod/i)%mod;
mul[1]=1;
for(int i=2;i<N;i++)
mul[i]=mul[i-1]*i%mod;
pt[1]=1;
for(int i=2;i<N;i++)
if(!vis[i])
pt[i]=pt[i-1]*(i-1)%mod*inv[i%mod]%mod;
else pt[i]=pt[i-1];
while(T--)
{
n=read();m=read();
printf("%lld\n",mul[n]*pt[m]%mod);
}
}//开个o2就过了扩欧板子题
cpp
#include<bits/stdc++.h>
using namespace std;
void gcd(int a,int b,int &x,int &y)
{
if(!b)
{
x=1;y=0;
return ;
}
gcd(b,a%b,x,y);
int xx=y,yy=x-a/b*y;
x=xx;y=yy;
return ;
}
int a,b,x,y;
int main()
{
scanf("%d%d",&a,&b);
gcd(a,b,x,y);
printf("%d",(x+b)%b);
}