POJ 3590 The shuffle Problem 置換+DP

題意：對每一個置換T，都存在一個T^k = e。現在讓你求一個n元置換，使得它的階最大，即當T^k = e時，k最大。若同時存在多個這樣的T，那麼輸出其中排序最小的。

題解：由於每一個置換都可以分解成若干個輪換，那麼這些輪換的階的最小公倍數就是該置換的階。

所以題目可以變成這樣：給你一個整數n，求n1+n2+n3```+ni = n。 並且n1,n2,```ni的最小公倍數最大。

1.求最小公倍數並不難，動態規劃解決。

2.那麼求得最小公倍數之後怎麼保證置換排序最小呢？

我們不妨令某個最小公倍數為lcmMax， 那麼將lcmMax因式分解之後得到 lcmMax = p1^k1*p2^k2*``pi^ki 

並且p1^k1+p2^k2+···+pi^ki <= n。這個是顯然的，因為 lcmMax = p1^k1*p2^k2*``pi^ki <= n1*n2*n3```*ni

而n1+n2+n3```+ni = n，所以p1^k1+p2^k2+···+pi^ki <= n。

3.用次用pi^ki個元素構成一個輪換，那麼就能保證該置換T的階最大。

那麼你可能會問，剩下的元素怎麼辦呢？其實全部讓它們為一階輪換就OK了，因為一階輪換並不影響最後T的階。

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define N 110#define lint __int64lint dp[N][N], maxLcm[N];lint factor[N], fnum;int p[25] ={2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31,37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97};inline lint gcd ( lint a, lint b ){    lint c;    while ( b != 0 )    {        c = a % b;        a = b;        b = c;    }    return a;}void split ( int n ){    int i, j, k, lcm;    memset(dp,0,sizeof(dp));    for ( i = 1; i <= n; i++ )        dp[i][1] = i;    for ( i = 2; i <= n; i++ )        for ( j = 2; j <= i; j++ )            for ( k = 1; k < i && i-k >= j-1; k++ )            {                lcm = dp[i-k][j-1] * k / gcd(dp[i-k][j-1], k);                if ( lcm > dp[i][j] ) dp[i][j] = lcm;            }    for ( i = 1; i <= n; i++ )    {        maxLcm[i] = 0;        for ( j = 1; j <= n; j++ )            if ( dp[i][j] >= maxLcm[i] )                maxLcm[i] = dp[i][j];    }}void split ( lint num ){    fnum = 0;    for ( int i = 0; i < 25; i++ )    {        if ( num % p[i] ) continue;        factor[fnum] = 1;        while ( num % p[i] == 0 )        {            factor[fnum] *= p[i];            num /= p[i];        }        fnum++;    }}int main(){    int t, n;    split(100);    scanf("%d",&t);    while ( t-- )    {        scanf("%d",&n);        split ( maxLcm[n] );        sort(factor,factor+fnum);        int i, j, k, tmp = 0;        for ( i = 0; i < fnum; i++ )            tmp += factor[i];        printf("%I64d",maxLcm[n]);        for ( i = 1; i <= n - tmp; i++ )            printf(" %d",i);        k = n - tmp;        for ( i = 0; i < fnum; i++ )        {            for ( j = 2; j <= factor[i]; j++ )                printf(" %d",k+j);            printf(" %d",k+1);            k += factor[i];        }        printf("\n");    }    return 0;}