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  • Train Problem II

    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. 

    InputThe 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. 
    OutputFor 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.
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<string>
    #include<set>
    #include<vector>
    #include<stack>
    #include<queue>
    #include<algorithm>
    #include<cstdio>
    #include<algorithm>
    #include<functional>
    #include<sstream>
    using namespace std;
    int a[105][105];    //大数卡特兰数
    int b[105];         //卡特兰数的长度
    
    void catalan()  //求卡特兰数
    {
        int i, j, len, carry, temp;
        a[1][0] = b[1] = 1;
        len = 1;
        for (i = 2; i <= 100; i++)
        {
            for (j = 0; j < len; j++)    //乘法
                a[i][j] = a[i - 1][j] * (4 * (i - 1) + 2);
            carry = 0;
            for (j = 0; j < len; j++)    //处理相乘结果
            {
                temp = a[i][j] + carry;
                a[i][j] = temp % 10;
                carry = temp / 10;
            }
            while (carry)    //进位处理
            {
                a[i][len++] = carry % 10;
                carry /= 10;
            }
            carry = 0;
            for (j = len - 1; j >= 0; j--) //除法
            {
                temp = carry * 10 + a[i][j];
                a[i][j] = temp / (i + 1);
                carry = temp % (i + 1);
            }
            while (!a[i][len - 1])     //高位零处理
                len--;
            b[i] = len;
        }
    }
    
    int main()
    {
        int i, n;
        catalan();
        while (scanf("%d", &n) != EOF)
        {
            for (i = b[n] - 1; i >= 0; i--)
            {
                printf("%d", a[n][i]);
            }
            printf("
    ");
        }
    
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/edych/p/7264006.html
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