[题解] [JLOI2013] 卡牌游戏

题面

题解

概率dp, 应该做得还是比较少的

设\(f[i][j]\)为该圈有\(i\)人时, 第\(j\)个人最后胜利的概率

枚举选择第几张卡牌, 设其值为\(card[k]\), 那么被淘汰的则是\(card[k] \% i\), 分类讨论

• \(card[k] \% i < j\), 则\(f[i][j] = f[i][j] + f[i - 1][(j - (card[k] \% i + 1)-1] / m\)
• \(card[k]\%i > j\), 则\(f[i][j] = f[i][j] + f[i - 1][i - ((card[k] \% i + 1 - j + 1) - 2)]\)
• \(card[k] \% i == j\), \(j\)被淘汰了...

就是这样, 最后算百分数不要忘记乘一百

Code

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
#define itn int
#define reaD read
#define N 55
using namespace std;

int n, m, card[N];
double f[N][N]; 

inline int read()
{
	int x = 0, w = 1; char c = getchar();
	while(c < '0' || c > '9') { if (c == '-') w = -1; c = getchar(); }
	while(c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar(); }
	return x * w;
}

int main()
{
	n = reaD(); m = read();
	for(int i = 1; i <= m; i++) card[i] = reaD();
	f[1][1] = 1;
	for(int i = 2; i <= n; i++)
		for(int j = 1; j <= i; j++)
			for(int k = 1; k <= m; k++)
			{
				int num = card[k] % i; if(!num) num = i;
				if(num > j) f[i][j] += f[i - 1][i - num + j] / m;
				else if(num < j) f[i][j] += f[i - 1][j - num] / m; 
			}
	for(int i = 1; i <= n; i++)
		printf("%.2lf%c%c", f[n][i] * 100, '%', i == n ? '\n' : ' '); 
	return 0; 
}