1023 Train Problem II_1023 ( train problem ii )-优快云博客

本文探讨了TrainProblemII的解决方法，通过卡特兰数的递推公式实现了大数运算，解决了火车按严格递增顺序进站后可能的出站顺序数量问题。提供了两种实现思路：一是直接存储预计算结果并输出；二是使用动态规划计算。

Train Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2034 Accepted Submission(s): 1191

Problem Description

As we all know the Train Problem I, the boss of the Ignatius Train Station want to know if all the trains come in strict-increasing order, how many orders that all the trains can get out of the railway.

Input

The input contains several test cases. Each test cases consists of a number N(1<=N<=100). The input is terminated by the end of file.

Output

For each test case, you should output how many ways that all the trains can get out of the railway.

Sample Input

Sample Output

1
2
5
16796

Hint
The result will be very large, so you may not process it by 32-bit integers.

Author

Ignatius.L

这一题确实有点小难，首先是卡特兰数，其次要大数相乘/除

#include <string>
#include <iostream>

using namespace std;

int main(int ac, char** av)
{
	string a[101] = {
	"0",
	"1",
	"2",
	"5",
	"14",
	"42",
	"132",
	"429",
	"1430",
	"4862",
	"16796",
	"58786",
	"208012",
	"742900",
	"2674440",
	"9694845",
	"35357670",
	"129644790",
	"477638700",
	"1767263190",
	"6564120420",
	"24466267020",
	"91482563640",
	"343059613650",
	"1289904147324",
	"4861946401452",
	"18367353072152",
	"69533550916004",
	"263747951750360",
	"1002242216651368",
	"3814986502092304",
	"14544636039226909",
	"55534064877048198",
	"212336130412243110",
	"812944042149730764",
	"3116285494907301262",
	"11959798385860453492",
	"45950804324621742364",
	"176733862787006701400",
	"680425371729975800390",
	"2622127042276492108820",
	"10113918591637898134020",
	"39044429911904443959240",
	"150853479205085351660700",
	"583300119592996693088040",
	"2257117854077248073253720",
	"8740328711533173390046320",
	"33868773757191046886429490",
	"131327898242169365477991900",
	"509552245179617138054608572",
	"1978261657756160653623774456",
	"7684785670514316385230816156",
	"29869166945772625950142417512",
	"116157871455782434250553845880",
	"451959718027953471447609509424",
	"1759414616608818870992479875972",
	"6852456927844873497549658464312",
	"26700952856774851904245220912664",
	"104088460289122304033498318812080",
	"405944995127576985730643443367112",
	"1583850964596120042686772779038896",
	"6182127958584855650487080847216336",
	"24139737743045626825711458546273312",
	"94295850558771979787935384946380125",
	"368479169875816659479009042713546950",
	"1440418573150919668872489894243865350",
	"5632681584560312734993915705849145100",
	"22033725021956517463358552614056949950",
	"86218923998960285726185640663701108500",
	"337485502510215975556783793455058624700",
	"1321422108420282270489942177190229544600",
	"5175569924646105559418940193995065716350",
	"20276890389709399862928998568254641025700",
	"79463489365077377841208237632349268884500",
	"311496878311103321137536291518809134027240",
	"1221395654430378811828760722007962130791020",
	"4790408930363303911328386208394864461024520",
	"18793142726809884575211361279087545193250040",
	"73745243611532458459690151854647329239335600",
	"289450081175264899454283846029490767264392230",
	"1136359577947336271931632877004667456667613940",
	"4462290049988320482463241297506133183499654740",
	"17526585015616776834735140517915655636396234280",
	"68854441132780194707888052034668647142985206100",
	"270557451039395118028642463289168566420671280440",
	"1063353702922273835973036658043476458723103404520",
	"4180080073556524734514695828170907458428751314320",
	"16435314834665426797069144960762886143367590394940",
	"64633260585762914370496637486146181462681535261000",
	"254224158304000796523953440778841647086547372026600",
	"1000134600800354781929399250536541864362461089950800",
	"3935312233584004685417853572763349509774031680023800",
	"15487357822491889407128326963778343232013931127835600",
	"60960876535340415751462563580829648891969728907438000",
	"239993345518077005168915776623476723006280827488229600",
	"944973797977428207852605870454939596837230758234904050",
	"3721443204405954385563870541379246659709506697378694300",
	"14657929356129575437016877846657032761712954950899755100",
	"57743358069601357782187700608042856334020731624756611000",
	"227508830794229349661819540395688853956041682601541047340",
	"896519947090131496687170070074100632420837521538745909320",	
	};

	int n;
	while( cin >> n ) {
		cout << a[n] << endl;
	}

	return 0;
}

呵呵，上面是个小玩笑，真正的解法代码在下面。下面的代码accepted后，突然想到这是确切的数值 1<=N<=100，把所有数存进去，然后打印出来不就行了……提交上面的代码后没想到还真能accepted这不知道算不算cheat(*^__^*) 嘻嘻……

/*
 *	Train Problem II
 *	2011/07/19	art
 */

/*
 *	卡特兰数
 *	h(n)= h(0)*h(n-1)+h(1)*h(n-2) + ... + h(n-1)h(0) (n>=2) 
 *	or
 *	h(n)=((4*n-2)/(n+1))*h(n-1);
 */

#include <stdio.h>

void init_a(int a[101][101]);
void printa(int a[101][101], int n);

int main(int ac, char** av)
{
	int a[101][101] = {0};	
	
	init_a(a);	// 初始化结果

/*	int n;
	while ( scanf("%d", &n) != EOF ) {
		printa(a, n);
	}
*/
	for ( int i = 1; i != 101; ++i ) {
		printa(a, i);
	}

	return 0;
}

void printa(int a[101][101], int n) {
	int j = a[n][0];

	if ( a[n][j] == 0 ) {
		;
	} else {
		printf("%d", a[n][j]);
	}
	--j;
	for ( ; j != 0; --j ) {
		printf("%04d", a[n][j]);	// 去除 0356 0056 之类情况	！
	}
	
	printf("\n");
}

void init_a(int a[101][101]) {
	a[1][1] = 1;
	a[2][1] = 2;
	a[1][0] = a[2][0] = 1;	// a[i][0] 存放数组长度
	
	int i = 0;

	for ( i = 3; i != 101; ++i) {
		int flag = 0;
		int temp = 0;
		int j = 0;
		
		for ( j = 1; j <= a[i-1][0]; ++j ) {	// 乘法部分 万位级
			// 可能会出现 0000 0354两类情况		！
			temp = a[i-1][j] * (4 * i - 2) + flag;
			a[i][j] = temp % 10000;
			flag = temp / 10000;
		}
		
		if ( flag != 0 ) {	// 检查最高位
			a[i][0] = j;
			a[i][j] = flag;
		} else {
			a[i][0] = j - 1;
		}
		
		
		flag = 0;
		for ( j = a[i][0]; j != 0; --j ) {	// 除法部分 万位级
			temp = a[i][j] + flag * 10000;
			a[i][j] = temp / (i + 1);
			flag = temp % (i + 1);
		}
		
		j = a[i][0];	// 去除 0000 此类情况		！
		if ( a[i][j] == 0 ) {
			--a[i][0];
		}
	}
}