数学问题的解题方法(模板)

查找数字规律公式的网站:  http://oeis.org/A005773

C++头文件:    #include <bits/stdc++>

链接:  http://blog.kuoe0.tw/posts/2014/01/31/install-gnu-gcc-on-os-x-and-use-the-header-bits-stdcplusplus-h-and-policy-based-data-structure/

一个数论讲的很好的博客:  http://www.cnblogs.com/linyujun/category/784324.html

辗转相除法:

1.最大公约数

int gcd(int a,int b){
    if(b==0)
        return a;
    return gcd(b,a%b);
}

LL gcd(LL a, LL b)
{
    return b == 0? a : gcd(b, a % b);
}

void ex_gcd(LL a, LL b, LL& d, LL& x, LL& y)
{
    if(!b)
    {
        d = a;
        x = 1; 
        y = 0;
    }
    else 
    {
        ex_gcd(b, a % b, d, y, x);
        y -= x * (a/b);
    }
}

2.扩展欧几里德算法

ax+by=gcd(a,b)

int extgcd(int a,int b,int &x,int &y){
    int d=a;
    if(b!=0){
        d=extgcd(b,a%b,y,x);
        y-=(a/b)*x;
    }
    else{
        x=1;
        y=0;
    }
    return d;
}

有关素数的基础算法:

1.素数判定

//素性测试
bool prime(int n){
    for(int i=2; i*i<=n; i++){
        if(n%i==0)
            return false;
    }
    return n!=1;  //1是例外
}

//约数枚举
vector<int> dis(int n){
    vector<int> res;
    for(int i=1; i*i<=n; i++){
        if(n%i==0){
            res.push_back(i);
            if(i!=n/i)
                res.push_back(n/i);
        }
    }
    return res;
}

//整数分解
map<int,int> prime(int n){
    for(int i=2; i*i<=n; i++){
        while(n%i==0){
            ++res[i];
            n/=i;
        }
    }
    if(n!=1)
        res[n]=1;
    return res;
}

2.埃氏筛法

int prime[MAX_N]; //第i个素数
bool is_prime[MAX_N+1]   //is_prime[i]为true表示i是素数

//返回n以内素数的个数
int solve(int n){
    int p=0;
    for(int i=0; i<=n; i++)
        is_prime[i]=true;
    is_prime[0]=is_prime[1]=false;
    for(int i=2; i<=n; i++){
        if(is_prime[i]){
            prime[p++]=i;
            for(int j=2*i; j<=n; j+=i)
                is_prime[i]=false;
        }
    }
    return p;
}

3. 区间筛法

typedef long long ll;

bool is_prime[MAX_L];
bool is_prime_small[MAX_B];

//对区间[a,b)内的整数执行筛法.is_prime[i-a] = true  --  i是素数
void segment_sieve(ll a,ll b){
    for(int i=0; (ll)i*i<b; i++)
        is_prime_small[i]=true;
    for(int i=0; i<b-a; i++)
        is_prime[i]=true;
    for(int i=2; (ll)i*i<b; i++){
        if(is_prime_small[i]){
            for(int j=2*i; (ll)j*j<b; j+=i)
                is_prime_small[j]=false;  //筛[2,根号b)
            for(ll j=max(2LL,(a+i-1)/i)*i; j<b; j+=i)
                is_prime[j-a]=false; //筛[a,b)
        }
    }
}

模运算

LL qpow(LL x, LL k)
{
    LL res = 1;
    while(k)
    {
        if(k & 1) res = res * x % MOD;
        x = x * x % MOD;
        k >>= 1;
    }
    return res;
}

LL inv(LL a, LL x)
{
    return qpow(a, x - 2);
}

凸包:

struct point
{
    double x;
    double y;
    point (double x = 0, double y = 0):x(x), y(y)
    {}
};
typedef point Vector;//向量
Vector operator + (Vector A, Vector B)//向量加法
{
    return Vector(A.x + B.x, A.y + B.y);
}

Vector operator - (Vector A, Vector B)//向量减法
{
    return Vector(A.x - B.x, A.y - B.y);
}

Vector operator * (Vector A, double p)//向量乘法
{
    return Vector(A.x * p, A.y * p);
}

Vector operator / (Vector A, double p)//向量除法
{
    return Vector(A.x / p, A.y / p);
}

int dcmp(double x)//精度控制
{
    if(fabs(x) < eps) return 0;
    else 
        return x < 0? -1 : 1;
}

bool operator == (const point& a, const point& b)//判断点相等
{
    return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}

double Dot(Vector A, Vector B)//向量点积，A*B，垂直为0
{
    return A.x * B.x + A.y * B.y;
}

double Length(Vector A)//向量长度
{
    return sqrt(Dot(A, A));
}

double Angle(Vector A, Vector B)//向量极角
{
    return acos(Dot(A, B) / Length(A) / Length(B));
}

double Cross(Vector A, Vector B)//向量叉乘，AXB， 有向面积的两倍
{
    return A.x * B.y - A.y * B.x;
}

double Area2(point A, point B, point C)//两向量组成的四边形的有向面积
{
    return Cross(B - A, C - A);
}
Vector Rotate(Vector A, double rad)//向量A，逆时针旋转rad°， 极角形式
{
    return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad));
}

Vector Normal(Vector A)//A will not 0 Vector!， 求单位向量
{
    double L = Length(A);
    return Vector(A.x / L, A.y / L);
}

Vector GetLineIntersection(point P, Vector v, point Q, Vector w)//求两向量的交点（不求具体坐标）
{
    Vector u = P - Q;
    double t = Cross(w, u) / Cross(v, w);
    return P + v * t;
}

double DistanceToLine(point P, point A, point B)// 点到直线的距离
{
    Vector v1 = B - A, v2 = P - A;
    return fabs(Cross(v1, v2)) / Length(v1);
}

double DistanceToSegment(point P, point A, point B)//点到线段的距离
{
    if(A == B) return Length(P - A);
    Vector v2 = P - A, v3 = P - B, v1 = B - A;
    if(dcmp(Dot(v1, v2) < 0)) return Length(v2);
    else if(dcmp(Dot(v1, v3)) > 0) return Length(v3);
    else 
        return fabs(Cross(v1, v2)) / Length(v1);
}

point GetLineProjection(point P, point A, point B)//求点在直线上的投影
{
    Vector v = B - A;
    return A + v * (Dot(v, P - A) / Dot(v, v));
}

bool SegmentProperIntersection(point a1, point a2, point b1, point b2)//两线段是否相交
{
    double c1 = Cross(a2 - a1, b1 - a1);
    double c2 = Cross(a2 - a2, b2 - a1);
    double c3 = Cross(b2 - b1, a1 - b1);
    double c4 = Cross(b2 - b1, a2 - b1);
    return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0;
}

bool OnSegment(point P, point a1, point a2)// 点是否在线段上
{
    return dcmp(Cross(a1 - P, a2 - P)) == 0 && dcmp(Dot(a1 - P, a2 - P)) < 0;
} 

double PolygonArea(point *p, int n)// 凸多边形（多边形）的有向面积
{
    double area = 0;
    for(int i = 1; i < n - 1;  i ++)
    area += Cross(p[i] - p[0], p[i + 1] - p[0]);
    return area / 2.0;
}

正多边形内接圆半径:

double getlen(double n,double r/){  //n正多边形边数,r正多边形边长
     return 2.0*r*tan(pi/n);   //const int PI = acos (-1.0);
}

 中国剩余定理模板:  http://blog.csdn.net/qq_32734731/article/details/51182391 

 

 数论可学习的链接:    http://www.cnblogs.com/linyujun/category/784324.html

 模板来自 : -> 萌哒哒~毅哥

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