数论基础

扩展欧几里得

hdu2669, 解同余方程 ax+by=1

int gcd(int a, int b, int& x, int& y)
{
    if(b == 0){
        x = 1;
        y = 0;
        return a;
    }
    else{
        int temp = gcd(b, a % b, x, y);
        int t = y;
        y = x - y * (a / b);
        x = t;
        return temp;
    }
}

费马小定理 gcd(a,p)=1，那么 a^(p-1) ≡1（mod p）

hdu4828,费马小定理计算卡特兰数

#include <iostream>
#include <cstdio>
#define MAXN 2000001
#define MOD 1000000007
using namespace std;

long long f[MAXN + 5];
long long gcd(long long a, long long b, long long& x, long long& y)
{
    if(b == 0){
        x = 1;
        y = 0;
        return a;
    }
    else{
        long long temp = gcd(b, a % b, x, y);
        long long t = y;
        y = x - y * (a / b);
        x = t;
        return temp;
    }
}

int main()
{
    f[1] = 1;
    for(long long i = 2; i <= MAXN; i++){
        f[i] = (f[i - 1] * (4 * i - 2)) % MOD;
        gcd(i + 1, MOD, x, y);
        if(x < 0){
            x = MOD - (-x) % MOD;
        }
        x = x % MOD;
        f[i] = (f[i] * x) % MOD;
    }

中国剩余定理

hdu1573,不互素的中国剩余定理

#include <iostream>
#include <cstdio>
#define M 11
using namespace std;

long long a[M], b[M];
long long x, y;
bool flag;
long long gcd(long long  a, long long  b)
{
    if (b == 0){
        return a;
    }
    return gcd(b, a % b);
}

long long exgcd(long long  a, long long  b, long long & x, long long & y)
{
    if (b == 0){
        x = 1;
        y = 0;
        return a;
    }
    else{
        long long  temp = exgcd(b, a % b, x, y);
        long long  t = y;
        y = x - y * (a / b);
        x = t;
        return temp;
    }
}
int main()
{
    long long  T, d, n, m, t, A, B;
    scanf("%I64d", &T);
    for (int cas = 1; cas <= T; cas++){
        flag = false;
        scanf("%I64d%I64d", &n, &m);
        for (int i = 1; i <= m; i++){
            scanf("%d", &a[i]);
            //t = (t * a[i]) / gcd(t, a[i]);
        }
        for (int i = 1; i <= m; i++){
            scanf("%I64d", &b[i]);
        }
        A = a[1], B = b[1];
        for (int i = 2; i <= m; i++){
            d = exgcd(A, a[i], x, y);
            if ((b[i] - B) % d != 0){
                flag = true;
                break;
            }
            x = (b[i] - B) / d * x;
            y = a[i] / d;
            x = (x % y + y) % y;
            B = x * A + B;
            A = (A * a[i]) / d;
            B = (B % A + A) % A;
        }
        if (B > n || flag){
            printf("0
");
        }
        else{
            t = 1 + (n - B) / A;
            if (B == 0){
                --t;
            }
            printf("%I64d
", t);
        }
    }
}

欧拉函数

求1-n与n互质的个数

int euler(int x){
     int ans = x, m = n;
     for(int i = 2; i * i <= x; i++){
         if(x % i == 0){
             ans = ans / i * (i - 1);
             while(x % i == 0){
                    x /= i;
             }
         }
     }
     if(x > 1){
            ans = ans / x * (x - 1);
     }
     return ans;
}

指数循环节