結果

問題 No.248 ミラー君の宿題
ユーザー anta
提出日時 2015-07-18 00:06:38
言語 C++11（廃止可能性あり）
(gcc 13.3.0)
結果
AC  
実行時間 482 ms / 5,000 ms
コード長 3,662 bytes
コンパイル時間 824 ms
コンパイル使用メモリ 90,176 KB
実行使用メモリ 5,376 KB
最終ジャッジ日時 2024-07-08 09:48:49
合計ジャッジ時間 4,211 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
other AC * 18
権限があれば一括ダウンロードができます
コンパイルメッセージ
main.cpp: In function ‘int main()’:
main.cpp:56:14: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   56 |         scanf("%d", &T);
      |         ~~~~~^~~~~~~~~~
main.cpp:59:22: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   59 |                 scanf("%d", &N);
      |                 ~~~~~^~~~~~~~~~
main.cpp:62:30: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   62 |                         scanf("%lld", &P[i]);
      |                         ~~~~~^~~~~~~~~~~~~~~

ソースコード

diff #
プレゼンテーションモードにする

#include <string>
#include <vector>
#include <algorithm>
#include <numeric>
#include <set>
#include <map>
#include <queue>
#include <iostream>
#include <sstream>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstring>
#include <cctype>
#include <cassert>
#include <limits>
#include <functional>
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define rer(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i))
#define reu(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i))
#if defined(_MSC_VER) || __cplusplus > 199711L
#define aut(r,v) auto r = (v)
#else
#define aut(r,v) __typeof(v) r = (v)
#endif
#define each(it,o) for(aut(it, (o).begin()); it != (o).end(); ++ it)
#define all(o) (o).begin(), (o).end()
#define pb(x) push_back(x)
#define mp(x,y) make_pair((x),(y))
#define mset(m,v) memset(m,v,sizeof(m))
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3fLL
using namespace std;
typedef vector<int> vi; typedef pair<int,int> pii; typedef vector<pair<int,int> > vpii; typedef long long ll;
template<typename T, typename U> inline void amin(T &x, U y) { if(y < x) x = y; }
template<typename T, typename U> inline void amax(T &x, U y) { if(x < y) x = y; }
//phi(P) | Q
//r^Q = 1
//p_i - 1 = 2^{e_i} × q_i
//p_i | h
//p_i | (c - 1)
//c ≡ 1 (mod p_i) 
//max_j {f_j(r)} = f_i(r)
//  { a^{2^k} = 1 | a <- S }
//= { 2^k×l ≡ 0 (mod p-1) | l <- [0..p-2] }
//= { v_2(l) >= v_2(p-1) - k | l <- [0..p-2] }
//= (p-1) / 2^{max(0, v_2(p-1) - k)}
//
int main() {
    int T;
    scanf("%d", &T);
    rep(ii, T) {
        int N;
        scanf("%d", &N);
        vector<long long> P(N);
        rep(i, N)
            scanf("%lld", &P[i]);
        vector<int> v_2(N);
        rep(i, N) {
            long long x = P[i] - 1;
            int k = 0;
            while(x % 2 == 0) x /= 2, ++ k;
            v_2[i] = k;
        }
        int maxv2 = *max_element(all(v_2));
        vector<double> dp(1 << N);
        dp[0] = 1;
        rep(i, N) dp[1 << i] = 1;
        static double dp2[1 << 13];
        static int masks[1 << 13];
        vi is;
        rep(s, 1 << N) if((s & (s-1)) != 0) {
            is.clear();
            rep(i, N) if(s >> i & 1)
                is.push_back(i);
            int n = is.size();
            rep(t, 1 << n) {
                int u = 0;
                rep(ix, n) if(t >> ix & 1)
                    u |= 1 << is[ix];
                masks[t] = u;
            }
            double multgroupfailprob = 0;
            double multgroupsum = 0;
            rer(k, 0, maxv2) {
                rep(t, 1 << n)
                    dp2[t] = 0;
                dp2[0] = 1;
                rep(ix, n) {
                    int v = v_2[is[ix]];
                    //1 / 2^{max(0, v_2[i] - k)}
                    double prob1 = 1. / (1LL << max(0, v - k));
                    double prob2 = k == 0 ? 0 : 1. / (1LL << max(0, v - (k - 1)));
                    rep(t, 1 << ix) {
                        double x = dp2[t];
                        if(x < 1e-99) continue;
                        dp2[t] = x * prob2;
                        dp2[t | 1 << ix] = x * (prob1 - prob2);
                    }
                }
                multgroupfailprob += dp2[(1 << n)-1];
                reu(t, 1, (1 << n)-1) {
                    double x = dp2[t];
                    if(x < 1e-99) continue;
                    int u = masks[t];
                    multgroupsum += x * (dp[u] + dp[s - u]);
                }
            }
            //dp2[t] = t が 0 になる確率
            rep(t, 1 << n)
                dp2[t] = 0;
            dp2[0] = 1;
            rep(ix, n) {
                long long p = P[is[ix]];
                double prob = 1. / p;
                rep(t, 1 << ix) {
                    double x = dp2[t];
                    if(x < 1e-99) continue;
                    dp2[t] = x * (1 - prob);
                    dp2[t | 1 << ix] = x * prob;
                }
            }
            double total = 0, failprob = 0;
            reu(t, 1, (1 << n)-1) {
                int u = masks[t];
                total += dp2[t] * (dp[u] + dp[s - u]);
            }
            failprob += dp2[(1 << n)-1];
            total += dp2[0] * multgroupsum;
            failprob += dp2[0] * multgroupfailprob;
            //E = 1 + failprob * E + total
            //E = (1 + total) / (1 - failprob)
            dp[s] = (1 + total) / (1 - failprob);
//          cerr << s << ": " << dp[s] << ", " << failprob << ", " <<total << endl;
        }
        double ans = dp[(1 << N)-1];
        printf("%.10f\n", ans);
    }
    return 0;
}
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