模板

1、线段相交

const double eps = 1e-8;

int sgn (double x) {
    if(fabs(x) < eps) return 0;
    if(x < 0) return -1;
    else return 1;
}

struct Point {
    double x, y;
    Point(){}
    Point(double xx, double yy): x(xx), y (yy) {}
    Point operator - (const Point &p) const {
        return Point(x - p.x, y - p.y);
    }
    double operator ^ (const Point  &p) const {     //判向量上侧、下侧  >0 P在上侧
        return x*p.y - y*p.x;
    }
    double operator * (const Point &p) const {   // 判向量夹角角度
        return x*p.x + y*p.y;
    }
};
struct Line{
    Point s, e;
    Line(){}
    Line(Point _s, Point _e): s(_s), e(_e){}
};

bool inter(Line l1, Line l2) {
    return max(l1.s.x, l1.e.x) >= min(l2.s.x, l2.e.x) && 
    max(l2.s.x, l2.e.x) >= min(l1.s.x, l1.e.x) && 
    max(l1.s.y, l1.e.y) >= min(l2.s.y, l2.e.y) &&
    

    max(l2.s.y, l2.e.y) >= min(l1.s.y, l1.e.y) && 
    sgn((l2.s - l1.s)^ (l1.e - l1.s)) * sgn((l2.e-l1.s) ^ (l1.e - l1.s)) <= 0 && 
    sgn((l1.s - l2.s)^ (l2.e - l2.s)) * sgn((l1.e-l2.s) ^ (l2.e - l2.s)) <= 0; 
}

2、素筛和唯一分解

const int maxn = 10100;

int prime[maxn];
bool vis[maxn];
int numm = 0;

void init_prime() {
    numm = 0;
    memset(prime, 0, sizeof(prime));
    memset(vis, true, sizeof(vis));
    for (int i = 2; i< maxn; i++) {
        if(vis[i]) {
            prime[numm] = i;
            numm += 1;
            for (int j = i*i; j < maxn; j += i) {
                vis[j] = false;
            }
            
        }
    }
    return ;
}

int s[maxn];
int cnt;

void solve(int a) {
    cnt = 0;
    for (int i = 0; prime[i] <= a; i++) {
        int pri = prime[i];
        if(a % pri == 0) {
            int num = 0;
            while(a % pri == 0) {
                num ++;
                a /= pri;
            }
            s[cnt ++] = pri;
        }
    }
}

int main() {
    init_prime();
    int a;
    while(cin >> a) {
        memset(s, 0, sizeof(s));
        solve(a);
        for (int i = 0; i < cnt; i++) {
            cout << s[i] << endl;
        }
    }
    system("pause");
    return 0;
}