P2059 [JLOI2013]卡牌游戏 概率DP

link:https://www.luogu.org/problemnew/show/P2059

题意：

有$n$个人，类似约瑟夫环的形式踢人，但是报的数是不同的，是在给定的许多数中随机抽取，问最后第$i$个人赢的概率；

思路： 

假设$dp[i][j]$，表示剩下$i$个人，从庄家后第$j$个人在本轮不被淘汰的概率。

首先枚举庄家抽到的卡牌$k$，得到这一轮被淘汰的人的位置$c$。如果$c==j$，就不要考虑了（因为这表示此轮第$j$个人被淘汰）。

而第$c$个人被淘汰之后，剩下的$i−1$个人要组成一个新的环，庄家为第$c$个人的下一个。容易算出，当$c>j$时，第$j$个人是新的环里从新庄家数起的第$i−c+j$个人，当$c<j$时，第$j$个人是新的环里从新庄家数起的第$j−c$个人。

 

#include <algorithm>
#include <iterator>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <stack>
#include <cmath>
#include <queue>
#include <list>
#include <map>
#include <set>
#include <cassert>

using namespace std;
#define lson (l, mid, rt << 1)
#define rson (mid + 1, r, rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue

typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
typedef pair<int, pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'

#define boost                    \
    ios::sync_with_stdio(false); \
    cin.tie(0)
#define rep(a, b, c) for (int a = (b); a <= (c); ++a)
#define max3(a, b, c) max(max(a, b), c);
#define min3(a, b, c) min(min(a, b), c);

const ll oo = 1ll << 17;
const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e9 + 7;
const double esp = 1e-8;
const double PI = acos(-1.0);
const double PHI = 0.61803399; //黄金分割点
const double tPHI = 0.38196601;

template <typename T>
inline T read(T &x) {
    x = 0;
    int f = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
    return x = f ? -x : x;
}

inline void cmax(int &x, int y) {
    if (x < y) x = y;
}
inline void cmax(ll &x, ll y) {
    if (x < y) x = y;
}
inline void cmin(int &x, int y) {
    if (x > y) x = y;
}
inline void cmin(ll &x, ll y) {
    if (x > y) x = y;
}

/*-----------------------showtime----------------------*/

const int maxn = 55;
int p[maxn];
double dp[maxn][maxn];
int main() {
    int n, m;
    scanf("%d%d", &n, &m);
    rep(i, 1, m) scanf("%d", &p[i]);
    dp[1][0] = 1.0;
    for (int i = 2; i <= n; i++) {
        //   for(int s = 0; s <i; s++){
        for (int j = 0; j < i; j++) {
            for (int t = 1; t <= m; t++) {
                int dx = (p[t] - 1) % i;
                if (dx == j) continue;
                if (dx > j) {
                    dp[i][j] += dp[i - 1][i - dx + j - 1] * 1.0 / m;

                } else
                    dp[i][j] += dp[i - 1][j - dx - 1] * 1.0 / m;
            }
        }
        // }
    }

    for (int i = 1; i <= n; i++) printf("%.2f%% ", 100 * dp[n][i - 1]);
    return 0;
}