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MZL's gameTime Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 261 Accepted Submission(s): 158 Problem Description MZL has $n$ cute boys.They are playing a game¡á.The game will run in turn First,System choose an alive player $x$ randomly.Player $x$ will be out of the game. Then player x will attack all alive players in the game When a player is attacked,$1-p$ is the probability of he still lives,$p$ is the probability of he dies Now mzl wants to know£ºthe probability of one player be out of the game and be attacked $k$ times You need to print the probability mod 258280327 for every k from 0 to n-1 According to Fermat Theory,$\frac{x}{y}$ mod 258280327=x*$(y^{258280325})$ mod 258280327 $p$ will be given in a special way Input The first line of the input contains a single number $T$, the number of test cases. Next $T$ lines, each line contains three integer $n$,$x$,$y$.$p=\frac{x}{y}$ $T\leq 5$, $n\leq 2*10^3$ $0 \leq x \leq 10^9$ $x+1 \leq y \leq 10^9$. It is guaranteed that y and 258280327 are coprime. Output $T$ lines, every line n numbers: the ans from 0 to n-1 Sample Input
Sample Output
Hint for case 1: The probability of you live and not be attacked is 1/3 The probability of you live and be attacked for one time is: (2/3)*(0.33*0.67+0.67*0.67*(1/2))=8911/30000 Author SXYZ Source | ||||||||||
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