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~ 跟：http://blog.csdn.net/qq_34374664/article/details/65935929對(duì)比理解下

?題目意思：有m個(gè)位置，每個(gè)位置填入一個(gè)數(shù)，數(shù)的范圍是1~n,問(wèn)至少有L個(gè)位置的數(shù)一樣的概率
?輸出要是最簡(jiǎn)分?jǐn)?shù)的形式，所以用大數(shù)JAVA

大致思路：

首先如果考慮有L, L + 1, .... m個(gè)位置上是一樣的方案數(shù)不好考慮, 但是可以從反面考慮, 計(jì)算只有1, 2, ... L - 1個(gè)位置有相同元素的方案數(shù), 用總方案數(shù)n^m減去即可

如果用dp[i][j]表示用前i種元素填了j個(gè)位置(不一定是前j個(gè)), 且每個(gè)元素都沒(méi)有使用達(dá)到L個(gè), 的方案數(shù)的話有

dp[i][j] = sigma(dp[i - 1][j - k]*C[m - (j - k)][k]) (0 <= k <= min(m, l))

那么最后的答案就是 (n^m-dp[1~n][m])/(n^m)

即用前i個(gè)元素填充j個(gè)位置的方案數(shù)相當(dāng)于前i - 1個(gè)元素填充了j - k個(gè)位置, 第i個(gè)元素要填充k個(gè)位置, 這k個(gè)位置可以來(lái)自于剩下的m - (j - k)個(gè)位置中的其中k個(gè)

那么不滿足題意的方案總數(shù)是 sigma(dp[i][m]) (1 <= i <= n) 用所有可能的排列方案減去即可得到可行的方案數(shù).

初始化dp[0][0] = 1, 其他值為0即可，這樣有些不符合的情況自動(dòng)變成0了

很明顯只有當(dāng)l > m時(shí)才會(huì)沒(méi)有可能的方案

考慮到計(jì)算時(shí)數(shù)據(jù)很大, 使用java方便一些

講道理 java真的是勁啊。。。?

import java.math.BigInteger; import java.util.BitSet; import java.util.Scanner;public class Main {static BigInteger[][] dp = new BigInteger[110][110];static BigInteger[][] c = new BigInteger[110][110];public static void main(String[] args){Scanner sc = new Scanner(System.in);for(int i = 0; i < 105; i++){c[i][0] = c[i][i] = BigInteger.valueOf(1);for(int j = 1; j < i; j++)c[i][j] = c[i-1][j-1].add(c[i-1][j]);}int n, m, l;while(sc.hasNext()){m = sc.nextInt();n = sc.nextInt();l = sc.nextInt();BigInteger tot = BigInteger.valueOf(n).pow(m);if(l > m){System.out.println("mukyu~");continue;} // if(l>m/2)//這個(gè)時(shí)候可以直接用組合數(shù)求出來(lái) // { // BigInteger ans=BigInteger.ZERO; // for(int i=l;i<=m;i++) // ans=ans.add(c[m][i].multiply(BigInteger.valueOf(n-1).pow(m-i))); // ans=ans.multiply(BigInteger.valueOf(n)); // BigInteger gcd=ans.gcd(tot); // System.out.println(ans.divide(gcd)+"/"+tot.divide(gcd)); // continue; // }for(int i = 0; i <= n; i++)for(int j = 0; j <= m; j++)dp[i][j] = BigInteger.valueOf(0);dp[0][0] = BigInteger.ONE; //0個(gè)元素房0個(gè)位置的概率肯定是1 啊for(int i = 1; i <= n; i++) //n個(gè)元素for(int j = 1; j <= m; j++) //j個(gè)位置，不一定連續(xù)for(int k = 0; k < l && k <= j; k++) //代表第i個(gè)元素不放，房1個(gè)，放兩個(gè)。。。最多放k個(gè)dp[i][j] = dp[i][j].add(dp[i-1][j-k].multiply(c[m-(j-k)][k])); // System.out.println(c[4][2]);BigInteger ans = BigInteger.ZERO;for(int i = 1; i <= n; i++)ans = ans.add(dp[i][m]);ans = tot.subtract(ans);BigInteger gcd = ans.gcd(tot);System.out.println(ans.divide(gcd)+"/"+tot.divide(gcd));}} }

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