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  • [题解] [JLOI2013] 卡牌游戏

    题面

    题解

    概率dp, 应该做得还是比较少的

    (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 ? '
    ' : ' '); 
    	return 0; 
    } 
    
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  • 原文地址:https://www.cnblogs.com/ztlztl/p/11206669.html
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