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  • [HDOJ1023]Train Problem II

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1023

     

    Train Problem II

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 6917    Accepted Submission(s): 3742


    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
    1 2 3 10
     
    Sample Output
    1 2 5 16796
    Hint
    The result will be very large, so you may not process it by 32-bit integers.
     
     
    ( 卡特兰数可以表示一个栈有多少个不同的出栈序列)
    大数卡特兰数的裸题,关于卡特兰数的介绍在这里:http://www.cppblog.com/MiYu/archive/2010/08/07/122573.html
     
    代码:
     
     1 #include <iostream>
     2 #include <stdio.h>
     3 #include <cmath>
     4 using namespace std;
     5 
     6 //Catalan
     7 int cat[105][105];    //大数卡特兰
     8 int clen[105];         //卡特兰数的长度
     9 int n;
    10 
    11 void catalan() {
    12     int i, j, len, carry, temp;
    13     cat[1][0] = clen[1] = 1;
    14     len = 1;
    15     for(i = 2; i <= 100; i++) {
    16         for(j = 0; j < len; j++)
    17             cat[i][j] = cat[i-1][j]*(4*(i-1)+2);
    18         carry = 0;
    19         for(j = 0; j < len; j++) {
    20             temp = cat[i][j] + carry;
    21             cat[i][j] = temp % 10;
    22             carry = temp / 10;
    23         }
    24         while(carry) {
    25             cat[i][len++] = carry % 10;
    26             carry /= 10;
    27         }
    28         carry = 0;
    29         for(j = len-1; j >= 0; j--) {
    30             temp = carry*10 + cat[i][j];
    31             cat[i][j] = temp/(i+1);
    32             carry = temp%(i+1);
    33         }
    34         while(!cat[i][len-1]) {
    35             len --;
    36         }
    37         clen[i] = len;
    38     }
    39 }
    40 
    41 void solve() {
    42     for(int i = clen[n] - 1; i >= 0; i--) {
    43         printf("%d", cat[n][i]);
    44     }
    45     printf("
    ");
    46 }
    47 
    48 int main() {
    49     catalan();
    50     while(scanf("%d", &n) != EOF) {
    51         solve();
    52     }
    53     return 0;
    54 }
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  • 原文地址:https://www.cnblogs.com/kirai/p/4753147.html
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