zoukankan      html  css  js  c++  java
  • HDU Train Problem II

    Train Problem II

    Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 138 Accepted Submission(s): 80
    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.

    #include"stdio.h"
    #include"string.h"
    int ch[101][200]={0};

    int temp[200]={0};

    int add(int* x,int* y)
    {
        int i;
        for(i=0;x[i]!=0&&y[i]!=0;i++)
        {
            x[i]+=y[i]-'0';
        }
        for(i;y[i]!=0;i++)
            x[i]+=y[i];
        int t;
        for(t=0;x[t]!=0;t++)
        {
            if(x[t+1]!=0)
                x[t+1]+=(x[t]-'0')/10;
            else
            {
                if((x[t]-'0')/10!=0)
                    x[t+1]+=(x[t]-'0')/10+'0';
                else
                    x[t+1]=0;
            }
            x[t]=(x[t]-'0')%10+'0';  
        }
        return t;
    }
    int* mul(int* x,int* y)
    {
        for(int i=0;i<200;i++)
            temp[i]=0;
      
        for(int i=0;x[i]!=0;i++)
            for(int j=0;y[j]!=0;j++)
            {
                if(temp[i+j]!=0)
                    temp[i+j]+=(x[i]-'0')*(y[j]-'0');
                else
                    temp[i+j]+=(x[i]-'0')*(y[j]-'0')+'0';
              
            }
        int length=0;
        while(temp[length]!=0)
            length++;
        for(int t=0;t<length;t++)
        {
            temp[t+1]+=(temp[t]-'0')/10;
            temp[t]=(temp[t]-'0')%10+'0';
        }
        if(temp[length]!=0)
            temp[length]+='0';
        while(temp[length]!=0)
        {
            if((temp[length]-'0')/10!=0)
                temp[length+1]+=temp[length]/10+'0';
            temp[length]=(temp[length]-'0')%10+'0';
            length++;
        }

        return temp;
    }

    int main()
    {
        ch[0][0]='1';
        ch[1][0]='1';

        for(int i=2;i<=100;i++)
        {
            for(int j=0;j<i;j++)
            {
                add(ch[i],mul(ch[j],ch[i-j-1]));
            }
        }
      
        int n,t=199;
        while(scanf("%d",&n)>0)
        {
            t=199;
            while(t>=0)
            {
                if(ch[n][t]!=0)
                    printf("%c",ch[n][t]);
                t--;
            }

            printf("\n");
        }
        return 0;
    }

  • 相关阅读:
    js showModalDialog参数传递
    1:N 关系
    1:N 关系 视图查找
    设置IE主页和添加收藏夹的功能
    GridView和DataFormatString网站技术
    JS 的table属性操作,
    GridView帮定数据显示数据的技巧
    后台取相同name值的问题
    赶集网CEO杨浩涌:倒闭没那么容易
    用户数据泄露案告破:嫌疑人已抓 CSDN受到警告
  • 原文地址:https://www.cnblogs.com/newpanderking/p/2121146.html
Copyright © 2011-2022 走看看