HDU 1865 1sting（大数斐波那契数列，模拟加法）

1sting

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4306    Accepted Submission(s): 1634

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

 

看懂题意以后，可以先推一下 111 和 1111的数量，不难推导出 111 的结果为 3 ，1111的结果为 4，因此可以得出，这道题是个递推题，递推公式就是斐波那契数列，但是，有一个难点就是要推出前 200 项的  ，这样  long long  的范围也肯定会超，所以就需要用到大数加法了

用二维数组，一维作为标记，二维作为储存

附上代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <string>

using namespace std;

int main()
{
	int s[250][50] = {0};         //  这里面我定义的数字为最长为50位，害怕不够的可以定义的大些，不过第 200 项的 数位是22 位
	s[1][0] = 1;
	s[2][0] = 2;
	s[3][0] = 3;
	for(int j = 4;j < 201;j++)
	{
		int jin = 0;
		for(int i = 0;i < 50;i++)
        {
            s[j][i] = s[j - 1][i] + s[j - 2][i] + jin;
            jin = 0;
            if(s[j][i] > 9 && i < 49)
            {
                jin = s[j][i]/10;
                s[j][i] = s[j][i]%10;
            }
        }
	}
	int t;
	cin >> t;
	while(t--)
	{
		char str[250];
		cin >> str;
		int flag = 0;
		for(int i = 49;i >= 0;i--)
		{
			if(s[strlen(str)][i] == 0 && flag == 0)           //  除去前面的 0
			{
				continue;
			}
			flag = 1;
			cout << s[strlen(str)][i];
		}
		cout << endl;
	}
	return 0;
}