HDU 1865 1sting （递推、大数） 1sting

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7573 Accepted Submission(s): 2945

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

Sample Output

1
2
8

题目大意

给你一个仅包含'1'的字符串; 你可以将两个相邻的“1”合并为“2”，或将“1”保留。例如，给定1111，可以获得1111,121,112,211,22。要求计算出可能的结果数量。

题目分析

当只有一个‘1’的时候，只有一种情况

　　有两个‘1’的时候，有两种情况，一种是 11 一种是 2

　　n>2的时候，我们可以在n-1的基础上在字符串的末尾加上一个 1

　　　　　　　　　　也可以在n-2的基础上加上一个2

所以得出递推公式：
　　f(n) = f(n-1) + f (n-2)

代码

#include<bits/stdc++.h>

using namespace std;

int n,i;
string x;
string bigadd(string a,string b)
{
    int jin=0,i;
    char ai,bi;
    string anss=a;
    int lena=a.size();
    int lenb=b.size();
    int lenmax=max(lena,lenb);
    int p=lena-1;
    int q=lenb-1;
    for(i=lenmax-1;i>=0;i--)
    {
        if(p<0)
        ai='0';
        else
        ai=a[p];
        if(q<0)
        bi='0';
        else
        bi=b[q];
        anss[i]=((ai-'0'+bi-'0'+jin)%10)+'0';
        jin=(ai-'0'+bi-'0'+jin)/10;
        p--;
        q--;
    }
    if(jin)
    {
        char x=jin+'0';
        anss=x+anss;
    }
    return anss;
}
 int main()
 {  
    string a[205];  
    a[1]="1";  
    a[2]="2";   
    for(i=3;i<205;++i)  
           a[i]=bigadd(a[i-1],a[i-2]);  
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    {
        cin>>x;
        cout<<a[x.size()]<<endl; 
    }
    return 0;       
}