2015 HNU warm up 07

A - Fractional Lotion

求有多少对(x, y)使得1/x + 1/y = 1/n。

(x+y) / (xy) = 1/n，设x=a*n，y=b*n，那么(a+b)/a*b=1。因为只有2^2 = 2*2，所以a和b两者必然分居2的两侧，即x和y必然一个大于2n，一个小于2n或者都等于2n。

#include <algorithm>#include <iostream>#include <stdlib.h>#include <string.h>#include <stdio.h>using namespace std;#define MAXN 10010int cnt[MAXN];void init(){    for(int i = 1; i < MAXN; i++)    {        cnt[i] = 0;        for(int j = i + 1; j <= 2 * i; j++) if((i * j) % (j - i) == 0) cnt[i]++;    }}int one, n, tot;int main(){//    freopen("A.in", "r", stdin);    init();    while(scanf("%d/%d", &one, &n) == 2)    {        printf("%d\n", cnt[n]);    }    return 0;}

D - Fence Orthogonality（最小周长外接矩形）

和UVa 12307类似，题解在这里。

#include <algorithm>#include <stdlib.h>#include <string.h>#include <iostream>#include <stdio.h>#include <math.h>using namespace std;#define MAXN 10010#define eps 1e-10template <class T>inline int RD(T &x){    x = 0;    char ch = getchar();    while(!isdigit(ch)) { ch = getchar(); if(ch == EOF) return 0; }    while(isdigit(ch)) { x *= 10; x += ch - '0'; ch = getchar(); }    return 1;}//const double pi = acos(-1.0);inline double sig(double x) { return (x > eps) - (x < -eps); };typedef struct Point{        double x, y;        Point() {}        Point(double _x, double _y):                x(_x), y(_y) {}        bool operator <(const Point &argu) const { return sig(x - argu.x) == 0 ? y < argu.y : x < argu.x; }        double dis(const Point &argu) const { return sqrt((x - argu.x) * (x - argu.x) + (y - argu.y) * (y - argu.y)); }        double dis2(const Point &argu) const { return (x - argu.x) * (x - argu.x) + (y - argu.y) * (y - argu.y); }        double operator ^(const Point &argu) const { return x * argu.y - y * argu.x; }        double operator *(const Point &argu) const { return x * argu.x + y * argu.y; }        Point operator -(const Point &argu) const { return Point(x - argu.x, y - argu.y); }        double len2() const { return x * x + y * y; }        double len() const { return sqrt(x * x + y * y); }        void in() { scanf("%lf%lf", &x, &y); }        void out() { printf("%.3lf %.3lf\n", x, y); }}Vector;inline double Cross(const Point &o, const Point &a, const Point &b) { return (a - o) ^ (b - o); }int ConvexHull(Point p[], Point ch[], int n){    sort(p, p + n);    int top = 0;    for(int i = 0; i < n; i++)    {        while(top > 1 && Cross(ch[top - 2], ch[top - 1], p[i]) <= 0) top--;        ch[top++] = p[i];    }    int t = top;    for(int i = n - 2; i >= 0; i--)    {        while(top > t && Cross(ch[top - 2], ch[top - 1], p[i]) <= 0) top--;        ch[top++] = p[i];    }    top--;    return top;}void RotatingCalipers(Point p[], int n, double &minp){    int t = 1, l = 1, r = 1;    minp = 1e15;    for(int i = 0; i < n; i++)    {        //枚举边(p[i], p[i+1])        while(sig((p[i + 1] - p[i]) ^ (p[t + 1] - p[t])) > 0) t = (t + 1) % n; //找出最高点        while(sig((p[i + 1] - p[i]) * (p[r + 1] - p[r])) > 0) r = (r + 1) % n; //找出最右点        if(i == 0) l = (r + 1) % n; //初始化最左点        while(sig((p[i + 1] - p[i]) * (p[l + 1] - p[l])) < 0) l = (l + 1) % n; //找出最左点        double d = p[i + 1].dis(p[i]);        double h = ((p[i + 1] - p[i]) ^ (p[t] - p[i])) / d; //三角形高        double w = ((p[i + 1] - p[i]) * (p[r] - p[l])) / d; //投影        minp = min(minp, 2 * (h + w));    }}Point pp[MAXN], ch[MAXN];int n, c;double minp;void solve(){    c = ConvexHull(pp, ch, n); ch[c] = ch[0];    RotatingCalipers(ch, c, minp);    printf("%.10lf\n", minp);}int main(){//    freopen("D.in", "r", stdin);    while(RD(n))    {        for(int i = 0; i < n; i++) RD(pp[i].x), RD(pp[i].y);        solve();    }    return 0;}