Patchouli's Spell Cards

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Time Limit: 7 Seconds Memory Limit: 65536 KB

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Patchouli Knowledge, the unmoving great library, is a magician who has settled down in the Scarlet Devil Mansion (紅魔館). Her specialty is elemental magic employing the seven elements fire, water, wood, metal, earth, sun, and moon. So she can cast different spell cards like Water Sign "Princess Undine", Moon Sign "Silent Selene" and Sun Sign "Royal Flare". In addition, she can combine the elements as well. So she can also cast high-level spell cards like Metal & Water Sign "Mercury Poison" and Fire, Water, Wood, Metal & Earth Sign "Philosopher's Stones" .

Assume that there are m different elements in total, each element has n different phase. Patchouli can use many different elements in a single spell card, as long as these elements have the same phases. The level of a spell card is determined by the number of different elements used in it. When Patchouli is going to have a fight, she will choose m different elements, each of which will have a random phase with the same probability. What's the probability that she can cast a spell card of which the level is no less than l, namely a spell card using at least l different elements.

Input

There are multiple cases. Each case contains three integers 1 ≤ m, n, l ≤ 100. Process to the end of file.

Output

For each case, output the probability as irreducible fraction. If it is impossible, output "mukyu~" instead.

Sample Input

7 6 5 7 7 7 7 8 9

Sample Output

187/15552 1/117649 mukyu~ 當(dāng)l>m明顯無解，l>m/2,可直接用組合數(shù)求出，當(dāng)l<=m/2,這里就不能用組合數(shù)求了，因?yàn)?#xff0c;這里會有重復(fù)，用dp[i][j]表示前i個數(shù)字，第j位，滿足某個數(shù)字小于l的個數(shù)，那么我們可以得出dp[i][j]=dp[i][j-k]c[m-j+k][k],也就是第i個數(shù)字，可以放１－k個位置，這樣就可以得出答案了！ import java.math.BigInteger; import java.util.Scanner; public class Main {public static void main(String[] args) {Scanner a=new Scanner(System.in);int M=150;BigInteger dp[][]=new BigInteger[150][150];BigInteger an[][]=new BigInteger[150][150];BigInteger one ,zero;one=BigInteger.ONE;zero=BigInteger.ZERO;for(int i=0;i<120;i++){dp[i][i]=dp[i][0]=dp[0][i]=one;for(int j=1;j<i;j++){dp[i][j]=dp[i-1][j-1].add(dp[i-1][j]);}}while(a.hasNext()){int m,n,l;BigInteger ans,nn,mm,ll,all,gcd;m=a.nextInt();n=a.nextInt();l=a.nextInt();nn=BigInteger.valueOf(n);mm=BigInteger.valueOf(m);ll=BigInteger.valueOf(l);ans=nn.pow(m);if(l>m){System.out.print("mukyu~\n");continue;}if(l>m/2){all=zero;for(int i=l;i<=m;i++){all=all.add(dp[m][i].multiply(BigInteger.valueOf(n-1).pow(m-i)));}all=BigInteger.valueOf(n).multiply(all);gcd=ans.gcd(all);System.out.println(all.divide(gcd)+"/"+ans.divide(gcd));continue;}for(int i=0;i<=n;i++)for(int j=0;j<=m;j++)an[i][j]=zero;an[0][0]=one; for(int i=1;i<=n;i++)for(int j=1;j<=m;j++){for(int k=0;k<=j&&k<l;k++){an[i][j]=an[i][j].add(an[i-1][j-k].multiply(dp[m-j+k][k]));}}all=zero;for(int i=1;i<=n;i++){all=all.add(an[i][m]);}all=ans.subtract(all);gcd=ans.gcd(all);System.out.println(all.divide(gcd)+"/"+ans.divide(gcd));}}}

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