数论模板

数论模板

就保存一下模板

P3811 【模板】乘法逆元

//递推
#include <cstdio>
#include <cstring>
#include <algorithm>

long long inv[3000010];

int main() {
	int n,p;
	scanf("%d%d",&n,&p);
	inv[1] = 1;
	printf("1
");
	for(int i = 2; i <= n; i++)
	{
		inv[i] = ((1ll * (-p / i) * inv[p % i] % p) + p) % p;
		printf("%d
",inv[i]);
	}
	return 0;
}

// 费马小定理 p为素数
#include <cstdio>
#include <algorithm>
#include <cstring>

using namespace std;

int poww(long long a,long long b)
{
	long long temp = 1,p = b-2;
	while(p)
	{
		if(p&1 == 1) temp = temp*a%b;
		a = a*a % b;
		p >>= 1;
	}
	return temp;
}

int main() {
	int n,p;
	scanf("%d%d",&n,&p);
	for(int i = 1; i <= n; i++)
	{
		long long ans = 0;
		ans = poww(i,p);
		printf("%lld
",ans);
	}
	return 0;
}

还有个exgcd

---

欧拉定理/扩展欧拉定理

【模板】欧拉定理

//单点求phi 模板题ac
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cctype>

using namespace std;

long long a, m, b, flag, ans = 1, phi, tmp;
char t;

void ksm(long long a, long long b, long long c) {
	a %= c;
	while(b) {
		if(b & 1) {
			ans = (ans * a) % c;
		}
		a = (a * a) % c;
		b >>= 1;
	}
}

int main() {
	scanf("%lld%lld", &a, &m);
	tmp = phi = m;
	for(long long i = 2; i * i <= m; i++) {
		if(tmp % i == 0) {
			phi -= (phi / i);
			while(tmp % i == 0) {
				tmp /= i;
			}
		}
	}
	if(tmp > 1) 
		phi -= (phi / tmp);
	while(!isdigit(t = getchar()));
	for(; isdigit(t); t = getchar()) {
		b = b * 10 + t - '0';
		if(b >= phi) {
			flag = 1;
			b %= phi;
		}
	}
	if(flag) {
		b += phi;
	}
	ksm(a, b, m);
	printf("%lld", ans);
	return 0;
} 

//递推求phi 模板题24分
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <string>
typedef int int_;
#define int long long

bool vis[1000007];
int pri[1000007], phi[1000007], cnt, temp, ans, lenm, lent, bns, m, a;
char strr[20000007];

inline void getphi(int n) {
	vis[0] = vis[1] = 1;
	phi[1] = 1;
	for(int i = 2; i <= n; i++)
	{
		if(vis[i] == 0)
		{
			pri[++cnt] = i;
			phi[i] = i - 1;
		}
		for(int j = 1; j <= cnt && pri[j] * i <= n; j++)
		{
			int next = pri[j] * i;
			vis[next] = 1;
			if(i % pri[j] == 0)
			{
				phi[next] = phi[i] * pri[j];
				break;
			}
			phi[next] = phi[i] * (pri[j] - 1);
		}
	}
}

inline int ksm(int a, int b, int mod)
{
	int ret = 1;
	while(b > 0)
	{
		if(b & 1) ret *= a, ret %= mod;
		a *= a;
		a %= mod;
		b >>= 1;
	}
	return ret;
}

int_ main() {
	scanf("%lld%lld", &a, &m);
	getphi(m);
	lenm = phi[m];
	while(lenm > 0)
	{
		lenm /= 10;
		lent++;
	}
	scanf("%s", strr);
	int len = strlen(strr), t = 1;
	for(int i = len - 1; i >= 0; i--)
	{
		temp = strr[i] - '0';
		ans += (t * temp);
		ans %= phi[m];
		t *= 10;
		t %= phi[m];
	}
	ans %= phi[m];
	bns = ksm(a, ans, m);
	if(ans == 0) bns = 0; 
	printf("%lld", bns);
	return 0;
}

---

P3390 【模板】矩阵快速幂

#include <cstdio>
#include <algorithm>
#include <cstring>
typedef int int_;           
#define int long long       
const int mod = 1e9 + 7;

struct MAT{
    int A[110][110];
    int x, y;
    MAT(int x_, int y_)
    {
        x = x_, y = y_;
        for(int i = 1; i <= x; i++)
            for(int j = 1; j <= y; j++)
                A[i][j] = 0;
    }
    void init(bool flg)
    {
        for(int i = 1; i <= x; i++)
            for(int j = 1; j <= y; j++)
                A[i][j] = 0;
        if(flg) 
            for(int i = 1; i <= x; i++) A[i][i] = 1;
    }
    MAT operator*(MAT b)
    {
        MAT a = *this;
        MAT c(a.x, b.y);
        c.init(0);
        for(int i = 1; i <= a.x; i++)
            for(int j = 1; j <= b.y; j++)
                for(int l = 1; l <= a.y; l++)
                {
                    c.A[i][j] += a.A[i][l] * b.A[l][j];
                    c.A[i][j] %= mod;
                }
        return c;
    }
    MAT operator^(int b)
    {
        MAT ret(x, x), a = *this;
        ret.init(1);
        while(b)
        {
            if(b&1) ret = ret * a;
            b >>= 1;
            a = a * a;
        }
        return ret;
    }
    void output()
    {
        for(int i = 1; i <= x; i++)
        {
            for(int j = 1; j <= y; j++)
                printf("%lld ", A[i][j]);
            puts("");
        }
    }
};

int_ main() {
    int n, m, k;
    scanf("%lld%lld", &n, &k);
    MAT a(n, n);
	
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++)
            scanf("%lld", &a.A[i][j]);
    MAT c = a ^ k;
    c.output();
}

---

P3807 【模板】卢卡斯定理

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
//I AK IOI
long long T, n, m, p;
long long jc[100007];

long long ksm(long long a, long long b, long long c) {
	a %= c;
	long long ret = 1;
	while(b != 0) {
		if(b & 1) ret = ret * a % c;
		b >>= 1;
		a = a * a % c;
	}
	return ret % c;
}

long long cm(long long a, long long b, long long c) {
	if(b > a) return 0;
	return (jc[a] * ksm(jc[b], c - 2, c) % c * ksm(jc[a - b], c - 2, c) % c) % c;
}

long long lucas(long long a, long long b, long long c) {
	if(b == 0) return 1;
	return cm(a % c, b % c, c) * lucas(a / c, b / c, c) % c;
} 

int main() {
	scanf("%lld", &T);
	jc[0] = 1;
	while(T--) {
		scanf("%lld%lld%lld", &n, &m, &p);
		for(long long i = 1; i <= p; i++) jc[i] = jc[i - 1] * i % p;
		printf("%lld
", lucas(m + n, n, p));
	}
} 

---

P1939 【模板】矩阵加速（数列）

#include <cstdio>
#include <algorithm>
#include <cstring>
typedef int int_;           
#define int long long       
const int mod = 1e9 + 7;

struct MAT{
    int A[110][110];
    int x, y;
    MAT(int x_, int y_)
    {
        x = x_, y = y_;
        for(int i = 1; i <= x; i++)
            for(int j = 1; j <= y; j++)
                A[i][j] = 0;
    }
    void init(bool flg)
    {
        for(int i = 1; i <= x; i++)
            for(int j = 1; j <= y; j++)
                A[i][j] = 0;
        if(flg) 
            for(int i = 1; i <= x; i++) A[i][i] = 1;
    }
    MAT operator*(MAT b)
    {
        MAT a = *this;
        MAT c(a.x, b.y);
        c.init(0);
        for(int i = 1; i <= a.x; i++)
            for(int j = 1; j <= b.y; j++)
                for(int l = 1; l <= a.y; l++)
                {
                    c.A[i][j] += a.A[i][l] * b.A[l][j];
                    c.A[i][j] %= mod;
                }
        return c;
    }
    MAT operator^(int b)
    {
        MAT ret(x, x), a = *this;
        ret.init(1);
        while(b)
        {
            if(b&1) ret = ret * a;
            b >>= 1;
            a = a * a;
        }
        return ret;
    }
};

int_ main() {
    int T;
    scanf("%lld", &T);
	while(T--)
	{
		int q;
		scanf("%lld", &q);
		if(q <= 3)
		{
			printf("1
");
			continue;
		}
		MAT ans(3, 3), chen(3, 1);
		ans.A[1][1] = ans.A[1][3] = ans.A[2][1] = ans.A[3][2] = 1;
		chen.A[1][1] = chen.A[2][1] = chen.A[3][1] = 1;
		ans = ans ^ (q - 3);
		ans = ans * chen;
    	printf("%lld
", ans.A[1][1]);
	}
	return 0;
}

---

P1226 【模板】快速幂||取余运算

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

int poww(long long a,long long b,long long c)
{
	long long temp = 1;
	while(b)
	{
		if(b&1 == 1) temp = temp*a%c;
		a = a * a % c;
		b >>= 1;
	}
	return temp;
}

int main() {
	long long b,p,k,ans;
	scanf("%lld%lld%lld",&b,&p,&k);
	ans = poww(b,p,k)%k;
	printf("%d^%d mod %d=%d",b,p,k,ans);
	return 0;
}

---

excrt

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

long long n, a[100007], p[100007];

long long ksc(long long x, long long y, long long mod) {
	long long ret = 0;
	while(y > 0) {
		if(y & 1) ret = (ret + x) % mod;
		y >>= 1;
		x = (x + x) % mod;
 	}
 	return (ret % mod + mod) % mod;
}

long long exgcd(long long m, long long n, long long &x, long long &y) {
	if(n == 0) {
		x = 1;
		y = 0;
		return m;
	}
	long long res = exgcd(n, m % n, y, x);
	y -= (m / n) * x;
	return res;
}

long long excrt() {
	long long A = a[1], P = p[1], x, y;
	for(long long i = 2; i <= n; i++) {
		long long a1 = P, b1 = p[i], c = ((a[i] - A) % b1 + b1) % b1;
		long long gcd = exgcd(a1, b1, x, y), bg = b1 / gcd;
		if(c % gcd != 0) return -1;//判无解
		x = ksc(x, c / gcd, bg);
		A += x * P;
		P *= bg;
		A = (A % P + P) % P;
	}
	if(A == 0) A = P;
	return A;
}

int main() {
//	freopen("111.in", "r", stdin);
	scanf("%lld", &n);
	for(long long i = 1; i <= n; i++) {
		scanf("%lld%lld", &p[i], &a[i]);
	}
	printf("%lld", excrt());
	return 0;
} 

---