Patchouli's Spell Cards

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=3957

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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~

References

題目大意：有m種不同的元素，每個(gè)元素有n種不同的階段，只有處于相同階段的元素才能放入咒語中，求一個(gè)咒語中至少有l(wèi)種不同的元素的概率？

題目大概直譯就是這樣，但是完全沒有思路

看了題解后發(fā)現(xiàn)大神總結(jié)抽象能力好強(qiáng)，抽象成：有m個(gè)位置，在每個(gè)位置隨機(jī)填上1~n個(gè)數(shù)，求相同的數(shù)至少有l(wèi)的概率？

即使這樣還是沒有思路，認(rèn)為是容斥什么的直接計(jì)算，完全想不到概率DP

大致思路如下：正難則反，統(tǒng)計(jì)填滿數(shù)后相同的數(shù)小于l的情況

設(shè)dp[i][j]表示前i個(gè)數(shù)占據(jù)j個(gè)位置的方案數(shù)（且每個(gè)數(shù)占據(jù)的位置小于l）

該狀態(tài)可由前i-1個(gè)數(shù)占據(jù)max(j-l+1,0) ~ j個(gè)位置轉(zhuǎn)移而來，保證第i個(gè)數(shù)占據(jù)的位置數(shù)小于l

則狀態(tài)轉(zhuǎn)移方程為：dp[i][j]=∑dp[i-1][j-k]*c(m-j+k,k) （k<=j&&k<l）

則∑dp[i][m] 表示用i個(gè)數(shù)填滿后，每個(gè)數(shù)出現(xiàn)次數(shù)都小于ll時(shí)的方案數(shù)

則n^m-∑dp[i][m] 為相同的數(shù)至少有l(wèi)的的方案數(shù)

import java.util.*; import java.math.*;public class Main {static BigInteger[][] dp=new BigInteger[105][105];static BigInteger[][] c=new BigInteger[105][105];static BigInteger fact,deno,gcd;public static void main(String[] argv) {for(int i=0;i<=100;++i) {c[i][0]=c[i][i]=BigInteger.ONE;for(int j=1;j<i;++j) {c[i][j]=c[i-1][j-1].add(c[i-1][j]);//楊輝三角形計(jì)算組合數(shù)}}Scanner cin=new Scanner(System.in);int m,n,l;while(cin.hasNext()) {m=cin.nextInt();n=cin.nextInt();l=cin.nextInt();if(l>m) {System.out.println("mukyu~");}else if(l>m/2) {//如果至少需要的階段相同的元素超過一半，則必定只有一種相同的階段fact=BigInteger.ZERO;deno=BigInteger.valueOf(n).pow(m);//合法的總方案數(shù)for(int i=l;i<=m;++i) {//枚舉階段相同的元素個(gè)數(shù)fact=fact.add(c[m][i].multiply(BigInteger.valueOf(n-1).pow(m-i)));//從m個(gè)元素中選擇i個(gè)元素其階段相同，其余m-i個(gè)元素可在剩下的n-1個(gè)階段任選}fact=fact.multiply(BigInteger.valueOf(n));//不同元素的相同階段共有n種gcd=fact.gcd(deno);//分子分母的最大公約數(shù)System.out.println(fact.divide(gcd)+"/"+deno.divide(gcd));//輸出最簡分?jǐn)?shù)}else {//存在多種相同的階段的元素個(gè)數(shù)都超過l時(shí)，進(jìn)行DPfor(int i=0;i<=n;++i) {for(int j=0;j<=m;++j) {dp[i][j]=BigInteger.ZERO;}}dp[0][0]=BigInteger.ONE;for(int i=1;i<=n;++i) {//枚舉階段for(int j=1;j<=m;++j) {//枚舉元素for(int k=0;k<l&&k<=j;++k) {//枚舉階段相同的元素dp[i][j]=dp[i][j].add(dp[i-1][j-k].multiply(c[m-j+k][k]));}}}fact=BigInteger.ZERO;deno=BigInteger.valueOf(n).pow(m);//合法的總方案數(shù)for(int i=1;i<=n;++i) {fact=fact.add(dp[i][m]);}fact=deno.subtract(fact);gcd=fact.gcd(deno);//分子分母的最大公約數(shù)System.out.println(fact.divide(gcd)+"/"+deno.divide(gcd));//輸出最簡分?jǐn)?shù)}}} }

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