Submission #302998

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Source Code Expand

#include <iostream>
#include <math.h>
#include <algorithm>
#include <string>
#include <stack>
#include <queue>
#include <set>
#include <cstdio>
#include <string.h>
using namespace std;

#define REP(i, n) for(int i = 0; i < n; i++)
#define FOR(i, b, e) for(int i = b; i < e; i++)
#define to_bit(i) static_cast< bitset<8> >(i)

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<long long, long long>PLL;
typedef queue<int> QI;
typedef priority_queue<int> maxpq;
typedef priority_queue<int, vector<int>, greater<int> > minpq;

struct edge{int to, cost;};

void begin(){cin.tie(0); ios::sync_with_stdio(false);};
int gcd(int a, int b){if(a%b==0){return(b);}else{return(gcd(b,a%b));}};
int lcm(int m, int n){if((0 == m)||(0 == n)){return 0;} return ((m / gcd(m, n)) * n);};
unsigned long long comb(long n, long m){unsigned long long c = 1; m = (n - m < m ? n - m : m);
    for(long ns = n - m + 1, ms = 1; ms <= m; ns ++, ms ++){c *= ns; c /= ms;} return c;};

void cp(int a1[], int a2[], int l){REP(i, l) a2[i] = a1[i];};
void cp(string a1[], string a2[], int l){REP(i, l) a2[i] = a1[i];};
double sq(double d){return d*d;};
int sq(int i){return i*i;};
double sqdist(int x1, int y1, int x2, int y2){
    double dx = x1 - x2, dy = y1 - y2; return dx*dx + dy*dy;
};
bool inside(int y, int x, int h, int w){return 0 <= y && y <= (h-1) && 0 <= x && x <= (w-1);};


// 線分の交差判定
bool isCross(int x1, int y1, int x2, int y2, int x3, int y3, int x4, int y4){
    // 並行な場合
    int m = (x2-x1)*(y4-y3)-(y2-y1)*(x4-x3);
    if(m == 0) return false;
    // 媒介変数の値が0より大きく1以下なら交差する、これは問題の状況によって変わるかも。
    double r =(double)((y4-y3)*(x3-x1)-(x4-x3)*(y3-y1))/m;
    double s =(double)((y2-y1)*(x3-x1)-(x2-x1)*(y3-y1))/m;
    return (0 < r && r <= 1 && 0 < s && s <= 1);
};

// 外積の計算 AB CD の内積を求める
int crossProdct(int ax, int ay, int bx, int by, int cx, int cy, int dx, int dy){
    int abx = bx - ax; int aby = by - ay;
    int cdx = dx - cx; int cdy = dy - cy;
    return abx * cdy - cdx * aby;
};
double crossProdct(double ax, double ay, double bx, double by, double cx, double cy, double dx, double dy){
    double abx = bx - ax; double aby = by - ay;
    double cdx = dx - cx; double cdy = dy - cy;
    return abx * cdy - cdx * aby;
};
double innerProduct(double ax, double ay, double bx, double by, double cx, double cy, double dx, double dy){
    double abx = bx - ax; double aby = by - ay;
    double cdx = dx - cx; double cdy = dy - cy;
    return abx * cdx + aby * cdy;
};

// 三角形の内部判定 ABCの中にPがあるか判定
bool inTriangle(int ax, int ay, int bx, int by, int cx, int cy, int px, int py){
    int c1 = crossProdct(ax, ay, bx, by, bx, by, px, py);
    int c2 = crossProdct(bx, by, cx, cy, cx, cy, px, py);
    int c3 = crossProdct(cx, cy, ax, ay, ax, ay, px, py);
    return (c1 > 0 && c2 > 0 && c3 > 0) || (c1 < 0 && c2 < 0 && c3 < 0);
};
bool inTriangle(double ax, double ay, double bx, double by, double cx, double cy, double px, double py){
    double c1 = crossProdct(ax, ay, bx, by, bx, by, px, py);
    double c2 = crossProdct(bx, by, cx, cy, cx, cy, px, py);
    double c3 = crossProdct(cx, cy, ax, ay, ax, ay, px, py);
    return (c1 > 0 && c2 > 0 && c3 > 0) || (c1 < 0 && c2 < 0 && c3 < 0);
};

// 三角形の外心
void circumcenter(double ax, double ay, double bx, double by, double cx, double cy, double res[3]){
    // AB AC を求める
    double abx = bx - ax; double aby = by - ay;
    double acx = cx - ax; double acy = cy - ay;
    double m = abx * acy - acx * aby; // 分母 m = 0 となるのは3点が一直線になるとき
    double s = (abx * acx + aby * acy - sq(acx) - sq(acy)) / m; // 媒介変数s
    res[0] = abx / 2 + s * aby / 2; res[1] = aby / 2 - s * abx / 2;
    // cout << res[0] << " " << res[1] << endl;
    res[2] = sqrt(sqdist(0, 0, res[0], res[1]));
    res[0] += ax; res[1] += ay;
};



class BinaryIndexedTree{
public:
    int n; vector<int> bit;
    BinaryIndexedTree(int _n){
        n = _n;
        bit.resize(n+1,0);
    }
    int sum(int i){
        int sum = 0;
        while(i > 0){
            sum += bit[i];
            i -= i & -i;
        }
        return sum;
    }
    void add(int i, int v){
        while(i <= n){
            bit[i] += v;
            i += i & -i;
        }
    }
};

class UnionFindTree{
public:
    vector<int> parent, rank;
    int n;
    std::set<int> set;
    // 初期化
    UnionFindTree(int _n){
        n = _n;
        for(int i = 0; i < n; i++){
            parent.resize(n);
            rank.resize(n);
            parent[i] = i;
            rank[i] = 0;
        }
    }
    // 根を求める
    int find(int x){
        if(parent[x] == x){
            return x;
        }else{
            return parent[x] = find(parent[x]);
        }
    }
    // xとyの集合を結合
    void unite(int x, int y){
        x = find(x);
        y = find(y);
        if(x == y){
            set.insert(x);
            return;
        }
        if(rank[x] < rank[y]){
            parent[x] = y;
            set.erase(x);
            set.insert(y);
        }else{
            parent[y] = x;
            set.erase(y);
            set.insert(x);
            if(rank[x] == rank[y]) rank[x]++;
        }
    }
    // xとyが同じ集合か
    bool same(int x, int y){
        return find(x) == find(y);
    }
    // 集合の数を数える
    int count(){
        return (int)set.size();
    }
};

// ワーシャルフロイド法
void warshallFloyd(int d[100][100], int n){
    for(int k = 0; k < n; ++k)
        for(int i = 0; i < n; ++i)
            for(int j = 0; j < n; ++j)
                if(d[i][k] != -1 && d[k][j] != -1){
                    if(d[i][j] == -1){
                        d[i][j] = d[i][k] + d[k][j];
                    }else{
                        d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
                    }
                }
};


// d:隣接行列  n:グラフのサイズ  s:始点  dist:始点からの距離が入る配列
void dijkstra(int d[1000][1000], int n, int s, int dist[1000]){
    REP(i, n) dist[i] = -1;
    dist[s] = 0;
    priority_queue<PII, vector<PII>, greater<PII> > q;
    q.push(PII(0, s));
    while (!q.empty()) {
        PII p = q.top(); q.pop();
        int i = p.second;
        if(dist[i] < p.first) continue;
        for(int j = 0; j < n; j++){
            if(d[i][j] == -1) continue;
            if(dist[j] == -1){
                dist[j] = dist[i] + d[i][j];
                q.push(PII(dist[j], j));
            }else if(dist[j] > dist[i] + d[i][j]){
                dist[j] = dist[i] + d[i][j];
                q.push(PII(dist[j], j));
            }
        }
    }
};


/**
 * start
 * @author yoshikyoto
 */

int n, coins[100];
int cou = 0;

int dfs(int d, int used[8], int perm[8]){
    // cout << "dfs " << d << endl;
    if(d == n){
        int flag[8]; REP(i, n) flag[i] = true;
        // REP(i, d) cout << perm[i] << " ";
        // cout << endl;
        REP(i, n){
            
            int val = perm[i];
            for(int j = i+1; j < n; j++){
                // cout << perm[j] << " " << val << endl;
                if(perm[j] % val == 0){
                    flag[j] = !flag[j];
                }
            }
        }
        int sum = 0;
        REP(i, n) sum += flag[i];
        // cout << sum << endl;
        cou++;
        return sum;
    }
    
    int sum = 0;
    REP(i, n){
        if(!used[i]){
            perm[d] = coins[i];
            used[i] = 1;
            sum += dfs(d+1, used, perm);
            used[i] = 0;
        }
    }
    return sum;
};

int main(int argc, const char * argv[]){
    begin();
    cin >> n;
    if(n > 8) return 0;
    REP(i, n) cin >> coins[i];
    
    int used[8]; REP(i, 8) used[i] = 0;
    int perm[8];
    // cout << dfs(0, used, perm) / (long double)cou  << endl;
    printf("%f", (dfs(0, used, perm) / (double)cou));
}

Submission Info

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