数学知识

试除法判断质数

#include <iostream>
#include <algorithm>

using namespace std;

bool is_prime(int x)
{
    if (x < 2) return false;
    for (int i = 2; i <= x / i; i ++ )
        if (x % i == 0)
            return false;
    return true;
}

int main()
{
    int n;
    cin >> n;

    while (n -- )
    {
        int x;
        cin >> x;
        if (is_prime(x)) puts("Yes");
        else puts("No");
    }

    return 0;
}

线性筛

#include <iostream>
#include <algorithm>

using namespace std;

const int N= 1000010;

int primes[N], cnt;
bool st[N];

void get_primes(int n)
{
    for (int i = 2; i <= n; i ++ )
    {
        if (!st[i]) primes[cnt ++ ] = i;
        for (int j = 0; primes[j] <= n / i; j ++ )
        {
            st[primes[j] * i] = true;
            if (i % primes[j] == 0) break;
        }
    }
}

int main()
{
    int n;
    cin >> n;

    get_primes(n);

    cout << cnt << endl;

    return 0;
}

约数个数

#include<iostream>
#include<unordered_map>
using namespace std;
const int mod=1e9+7;
int main()
{
    int n;
    cin>>n;
    unordered_map<int,int>mp;
    while(n--)
    {
        int x;
        cin>>x;
        for(int i=2;i<=x/i;i++)
        {
            if(x%i==0)
            {
                while(x%i==0)
                    mp[i]++,x/=i;
            }
        }
        if(x>1) mp[x]++;
    }
    long long res=1;
    for(auto x:mp)
    {
        int a=x.first,b=x.second;
        res=res*(1+b)%mod;
    }
    cout<<res%mod<<endl;
}

欧拉函数

代码

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define mst(s,_s) memset(s, _s, sizeof(s))
const double PI = acos(-1.0);
const double eps = 1e-6;
const int INF = 0x3f3f3f3f;
const int N = 1e6+100;
int T,n,m;




int main() {

    int n;
    cin>>n;
    while(n--)
    {
        int a;
        cin>>a;
        int res=a;
        for(int i=2;i<=a/i;i++)
        {
            if(a%i==0)
            {
                res = res/i*(i-1);
                while(a%i==0)
                    a/=i;
            }
        }
        if(a>1) res=res/a*(a-1);
        cout<<res<<endl;
    }
    return 0;
}

筛法求欧拉函数

代码

#include<bits/stdc++.h>
using namespace std;
const int N=2e6;
typedef long long ll;
ll phi[N],res;
int st[N],prime[N],cnt;
void init(int n)
{
    phi[1]=1;
    for(int i=2;i<=n;i++)
    {
        if(!st[i]){
            prime[cnt++]=i;
            phi[i]=i-1;
        }
        for(int j=0;prime[j]<=n/i;j++)
        {
            st[prime[j]*i]=1;
            if(i%prime[j]==0)
            {
                phi[prime[j]*i] =phi[i] *prime[j];
                break;
            }
            phi[prime[j]*i] = phi[i]*(prime[j]-1);
        }
    }
}
int main()
{
    int n;
    cin>>n;
    init(n);
    ll res=0;
    //cout<<"n=="<<n<<endl;
   for(int i=1;i<=n;i++) res+=phi[i];
   cout<<res<<endl;
    //cout<<res<<endl;
}

快速幂

#include <iostream>
#include <algorithm>

using namespace std;

typedef long long LL;


LL qmi(int a, int b, int p)
{
    LL res = 1 % p;
    while (b)
    {
        if (b & 1) res = res * a % p;
        a = a * (LL)a % p;
        b >>= 1;
    }
    return res;
}


int main()
{
    int n;
    scanf("%d", &n);
    while (n -- )
    {
        int a, b, p;
        scanf("%d%d%d", &a, &b, &p);
        printf("%lld\n", qmi(a, b, p));
    }

    return 0;
}

求组合数一（a,b,1000)

代码（杨辉三角

#include <iostream>
#include <algorithm>

using namespace std;

const int N = 2010, mod = 1e9 + 7;


int c[N][N];


void init()
{
    for (int i = 0; i < N; i ++ )
        for (int j = 0; j <= i; j ++ )
            if (!j) c[i][j] = 1;
            else c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
}


int main()
{
    int n;

    init();

    scanf("%d", &n);

    while (n -- )
    {
        int a, b;
        scanf("%d%d", &a, &b);

        printf("%d\n", c[a][b]);
    }

    return 0;
}

求组合数二(a,b,10000)

代码

#include <iostream>
#include <algorithm>

using namespace std;

typedef long long LL;

const int N = 100010, mod = 1e9 + 7;


int fact[N], infact[N];


int qmi(int a, int k, int p)
{
    int res = 1;
    while (k)
    {
        if (k & 1) res = (LL)res * a % p;
        a = (LL)a * a % p;
        k >>= 1;
    }
    return res;
}


int main()
{
    fact[0] = infact[0] = 1;
    for (int i = 1; i < N; i ++ )
    {
        fact[i] = (LL)fact[i - 1] * i % mod;
        infact[i] = (LL)infact[i - 1] * qmi(i, mod - 2, mod) % mod;
    }


    int n;
    scanf("%d", &n);
    while (n -- )
    {
        int a, b;
        scanf("%d%d", &a, &b);
        printf("%d\n", (LL)fact[a] * infact[b] % mod * infact[a - b] % mod);
    }

    return 0;
}

求组合数三（a,b超级大）

代码:卢卡斯定理

#include <iostream>
#include <algorithm>

using namespace std;

typedef long long LL;


int qmi(int a, int k, int p)
{
    int res = 1;
    while (k)
    {
        if (k & 1) res = (LL)res * a % p;
        a = (LL)a * a % p;
        k >>= 1;
    }
    return res;
}


int C(int a, int b, int p)
{
    if (b > a) return 0;

    int res = 1;
    for (int i = 1, j = a; i <= b; i ++, j -- )
    {
        res = (LL)res * j % p;
        res = (LL)res * qmi(i, p - 2, p) % p;
    }
    return res;
}


int lucas(LL a, LL b, int p)
{
    if (a < p && b < p) return C(a, b, p);
    return (LL)C(a % p, b % p, p) * lucas(a / p, b / p, p) % p;
}


int main()
{
    int n;
    cin >> n;

    while (n -- )
    {
        LL a, b;
        int p;
        cin >> a >> b >> p;
        cout << lucas(a, b, p) << endl;
    }

    return 0;
}