題目鏈接：https://vjudge.net/problem/HDU-2067

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小兔的棋盤

Time Limit: 1000/1000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11800????Accepted Submission(s): 5952

Problem Description 小兔的叔叔從外面旅游回來給她帶來了一個(gè)禮物，小兔高興地跑回自己的房間，拆開一看是一個(gè)棋盤，小兔有所失望。不過沒過幾天發(fā)現(xiàn)了棋盤的好玩之處。從起點(diǎn)(0，0)走到終點(diǎn)(n,n)的最短路徑數(shù)是C(2n,n),現(xiàn)在小兔又想如果不穿越對(duì)角線(但可接觸對(duì)角線上的格點(diǎn))，這樣的路徑數(shù)有多少?小兔想了很長(zhǎng)時(shí)間都沒想出來，現(xiàn)在想請(qǐng)你幫助小兔解決這個(gè)問題，對(duì)于你來說應(yīng)該不難吧! Input 每次輸入一個(gè)數(shù)n(1<=n<=35)，當(dāng)n等于－1時(shí)結(jié)束輸入。 Output 對(duì)于每個(gè)輸入數(shù)據(jù)輸出路徑數(shù)，具體格式看Sample。 Sample Input 1 3 12 -1 Sample Output 1 1 2 2 3 10 3 12 416024 Author Rabbit Source RPG專場(chǎng)練習(xí)賽 Recommend lcy

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題解：

1?卡特蘭數(shù)的初步學(xué)習(xí)，卡特蘭數(shù)應(yīng)用 。

2.卡特蘭數(shù)計(jì)算公式：

?1） h(n) = h(0)*h(n-1) + h(1)*h(n-2) + ... + h(n-1)h(0) (n>=1) ， 其中 h[0] = 1；

?2） h(n) = c(2n,n) - c(2n,n+1)(n=0,1,2,...) <==>?h(n) = C(2n,n)/(n+1)

?3） h(n) = h(n-1)*(4*n-2) / (i+1)? ……此條計(jì)算公式容易溢出

注意：卡特蘭數(shù)的計(jì)算很容易溢出。

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代碼如下：

1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <vector> 6 #include <cmath> 7 #include <queue> 8 #include <stack> 9 #include <map> 10 #include <string> 11 #include <set> 12 using namespace std; 13 typedef long long LL; 14 const int INF = 2e9; 15 const LL LNF = 9e18; 16 const int MOD = 1e9+7; 17 const int MAXN = 35+10; 18 19 LL h[MAXN]; 20 21 void init() 22 { 23 memset(h, 0, sizeof(h)); 24 h[0] = 1; h[1] = 1; 25 for(int i = 2; i<MAXN; i++) 26 for(int j = 0; j<i; j++) 27 h[i] += 1LL*h[j]*h[i-j-1]; 28 } 29 30 int main() 31 { 32 init(); 33 int kase = 0, n; 34 while(scanf("%d", &n) && n!=-1) 35 printf("%d %d %lld\n", ++kase, n, 2LL*h[n]); 36 } View Code

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題目鏈接：?https://vjudge.net/problem/HDU-4165

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Pills

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1626????Accepted Submission(s): 1139

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Problem Description Aunt Lizzie takes half a pill of a certain medicine every day. She starts with a bottle that contains N pills.

On the first day, she removes a random pill, breaks it in two halves, takes one half and puts the other half back into the bottle.

On subsequent days, she removes a random piece (which can be either a whole pill or half a pill) from the bottle. If it is half a pill, she takes it. If it is a whole pill, she takes one half and puts the other half back into the bottle.

In how many ways can she empty the bottle? We represent the sequence of pills removed from the bottle in the course of 2N days as a string, where the i-th character is W if a whole pill was chosen on the i-th day, and H if a half pill was chosen (0 <= i < 2N). How many different valid strings are there that empty the bottle? Input The input will contain data for at most 1000 problem instances. For each problem instance there will be one line of input: a positive integer N <= 30, the number of pills initially in the bottle. End of input will be indicated by 0. Output For each problem instance, the output will be a single number, displayed at the beginning of a new line. It will be the number of different ways the bottle can be emptied. Sample Input 6 1 4 2 3 30 0 Sample Output 132 1 14 2 5 3814986502092304 Source The 2011 Rocky Mountain Regional Contest

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Recommend lcy

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題解：

有n片藥，每天吃半片。當(dāng)天要么在藥罐中抽到把一片完整的藥片，然后分成兩半，吃一半，最后把另一半放回藥罐中；要么抽到半片藥片直接吃。問：有多少種情況？ 單純的卡特蘭數(shù)。

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代碼如下：

1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <vector> 6 #include <cmath> 7 #include <queue> 8 #include <stack> 9 #include <map> 10 #include <string> 11 #include <set> 12 using namespace std; 13 typedef long long LL; 14 const int INF = 2e9; 15 const LL LNF = 9e18; 16 const int MOD = 1e9+7; 17 const int MAXN = 30+10; 18 19 LL h[MAXN]; 20 21 void init() 22 { 23 memset(h, 0, sizeof(h)); 24 h[0] = 1; 25 for(int i = 1; i<MAXN; i++) 26 for(int j = 0; j<i; j++) 27 h[i] += 1LL*h[j]*h[i-j-1]; 28 } 29 30 int main() 31 { 32 init(); 33 int n; 34 while(scanf("%d", &n) && n) 35 printf("%lld\n", h[n]); 36 } View Code

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題目鏈接：https://vjudge.net/problem/HDU-1134

Game of Connections

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5112????Accepted Submission(s): 2934

Problem Description This is a small but ancient game. You are supposed to write down the numbers 1, 2, 3, ... , 2n - 1, 2n consecutively in clockwise order on the ground to form a circle, and then, to draw some straight line segments to connect them into number pairs. Every number must be connected to exactly one another. And, no two segments are allowed to intersect.

It's still a simple game, isn't it? But after you've written down the 2n numbers, can you tell me in how many different ways can you connect the numbers into pairs? Life is harder, right? Input Each line of the input file will be a single positive number n, except the last line, which is a number -1. You may assume that 1 <= n <= 100. Output For each n, print in a single line the number of ways to connect the 2n numbers into pairs. Sample Input 2 3 -1

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Sample Output 2 5

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Source Asia 2004, Shanghai (Mainland China), Preliminary Recommend Eddy

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題意：

1~2*n 順時(shí)針排列成一圈， 用n條線段連接n對(duì)數(shù)，要求線段不能有交叉，問：有多少種連接情況？

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題解：

可以將此題聯(lián)想到出棧問題，這樣就轉(zhuǎn)化成卡特蘭數(shù)了。

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遞推式一：

1 import java.util.Scanner; 2 import java.math.BigInteger; 3 4 public class Main { 5 6 public static void main(String[] args){ 7 8 BigInteger[] a = new BigInteger[105]; 9 10 a[0] = BigInteger.ONE; 11 for(int i=1; i<=100; i++) { 12 a[i] = BigInteger.valueOf(0); 13 for(int j=0;j<i;j++){ 14 a[i] = a[i].add(a[j].multiply(a[i-j-1])); 15 } 16 } 17 18 Scanner input = new Scanner(System.in); 19 while(input.hasNext()){ 20 int n=input.nextInt(); 21 if(n==-1) break; 22 System.out.println(a[n]); 23 } 24 } 25 } View Code

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遞推式二：

1 import java.util.Scanner; 2 import java.math.BigInteger; 3 4 public class Main { 5 6 public static void main(String[] args){ 7 8 BigInteger[] a = new BigInteger[105]; 9 10 a[0] = BigInteger.ONE; 11 for(int i=1; i<=100; i++) { 12 a[i] = a[i-1].multiply(BigInteger.valueOf(4*i-2)).divide(BigInteger.valueOf(i+1)); 13 } 14 15 Scanner input = new Scanner(System.in); 16 while(input.hasNext()){ 17 int n=input.nextInt(); 18 if(n==-1) break; 19 System.out.println(a[n]); 20 } 21 } 22 } View Code

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轉(zhuǎn)載于:https://www.cnblogs.com/DOLFAMINGO/p/8324436.html

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