HDOJ 1865 1sting 13.04.21 周赛结题报告 大数加法

1sting

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2227 Accepted Submission(s): 875

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
  
  
  
  


  
  
  
  

   
   解题思路：
  
  
  
  

   
    一开始还以为要用组合数学来做了。。
  
  
  
  

   
    后来发现原来是个斐波那契数列。
  
  
  
  

   
    悲剧的是将200的字符串长度看成了20。。。 硬生生的把一道高精度加法当做了简单加法做。。
  
  
  
  

   
   算法：
  
  
  
  

   
    高精度计算。
  
  
  
  

   
   代码：
  
  
  
  

   
   
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int a[201][101]={0};
int main ()
{
    int n,i,j;
    a[1][100]=1;
    a[2][100]=2;
    for (i=3; i<=200; i++)
    {
        for (j=100; j>=0; j--)
        {
            a[i][j]+=a[i-1][j]+a[i-2][j];
            if (a[i][j]>=10)
            {
                a[i][j-1]+=a[i][j]/10;
                a[i][j]=a[i][j]%10;
            }
        }
    }
    cin>>n;
    while(n--)
    {
        char str[210];
        cin>>str;
        int l=strlen(str);
        int p;
        for (j=0; j<=100; j++)
            if (a[l][j]!=0)
                break;
        p=j;
        for (j=p; j<=100; j++)
            cout<<a[l][j];
        cout<<endl;
    }
    return 0;
}