zoukankan      html  css  js  c++  java
  • HDUOJ--1865 1string

    1sting

    Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 2372    Accepted Submission(s): 947


    Problem Description
    You will be given a string which only contains ‘1’; You can merge two adjacent ‘1’ to be ‘2’, or leave the ‘1’ there. Surly, you may get many different results. For example, given 1111 , you can get 1111, 121, 112,211,22. Now, your work is to find the total number of result you can get.
     
    Input
    The first line is a number n refers to the number of test cases. Then n lines follows, each line has a string made up of ‘1’ . The maximum length of the sequence is 200.
     
    Output
    The output contain n lines, each line output the number of result you can get .
     
    Sample Input
    3
    1
    11
    11111
     
    Sample Output
    1
    2
    8
     
    Author
    z.jt
     
    Source
    Recommend
    lcy
     简单大数题....
    代码:
     1 #include<iostream>
     2 #include<string>
     3 #define maxn 50
     4 #define  len 200
     5 using namespace std;
     6 int arr[len][maxn+1]={{1},{1}};
     7 int main()
     8 {
     9     int i,j,s,c,t;
    10     string n ;
    11     for(i=2;i<=len;i++)
    12     {
    13         for(c=j=0;j<=maxn;j++)
    14         {
    15           s=arr[i-1][j]+arr[i-2][j]+c;
    16           arr[i][j]=s%10;
    17           c=(s-arr[i][j])/10;
    18         }
    19     }
    20     cin>>t;
    21     while(t--)
    22     {
    23         cin>>n;
    24         for(i=maxn;arr[n.size()][i]==0;i--);
    25         for(j=i;j>=0;j--)
    26         {
    27             printf("%d",arr[n.size()][j]);
    28         }
    29         puts("");
    30     }
    31 return 0;
    32 }
    View Code
  • 相关阅读:
    SQL 函数以及SQL 编程
    查询练习(四十五道题)
    SQL server 高级查询
    SQL server 数据库 操作及简单查询
    T-SQL 语句
    SQL
    表单验证、实则表达式、事件
    视频插入代码
    插入地图
    网页布局小练
  • 原文地址:https://www.cnblogs.com/gongxijun/p/3236426.html
Copyright © 2011-2022 走看看