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2020-2021:teams:wangzai_milk:wzx27:combinatorial_mathematics:permutaitiongroup [CVBB ACM Team]

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两侧同时换到之前的修订记录 前一修订版
2020-2021:teams:wangzai_milk:wzx27:combinatorial_mathematics:permutaitiongroup [2020/05/29 02:33]
wzx27
2020-2021:teams:wangzai_milk:wzx27:combinatorial_mathematics:permutaitiongroup [2020/05/29 02:34] (当前版本)
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 有上述定理可知$T^k=\iota$的最小正整数解为$T$中所有循环的最小公倍数,这个显然能通过$\text{dp}$求得。记$f[i][j]:​=和为i的j个数的lcm$,于是$f[i][j]=max\{f[i-1][j-k]*k/​gcd(f[i-1][j-k],​k)\}$ 有上述定理可知$T^k=\iota$的最小正整数解为$T$中所有循环的最小公倍数,这个显然能通过$\text{dp}$求得。记$f[i][j]:​=和为i的j个数的lcm$,于是$f[i][j]=max\{f[i-1][j-k]*k/​gcd(f[i-1][j-k],​k)\}$
  
-从而能得到长度为$n$的置换的解并唯一分解得到$maxlcm=\prod p_i^{k_i}$且$\sum p_i^{k_i}\le n$,所以拆成大小为$p_i^{k_i}$的循环,剩下的都是长度为1的循环即可+从而能得到长度为$n$的置换的解$\text{maxlcm}$,​并唯一分解得到$maxlcm=\prod p_i^{k_i}$且$\sum p_i^{k_i}\le n$,所以拆成大小为$p_i^{k_i}$的循环,剩下的都是长度为1的循环即可
  
 2、置换群的幂运算 2、置换群的幂运算
2020-2021/teams/wangzai_milk/wzx27/combinatorial_mathematics/permutaitiongroup.1590690809.txt.gz · 最后更改: 2020/05/29 02:33 由 wzx27