A专题-05递推与组合 --- C - 1sting

C - 1sting

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

题记：用记忆化递归求斐波那契数列，可以用高精度加法计算。

#include<iostream>
#include<vector>
#include<string>
using namespace std;
string add(string a, string b)
{
	vector <int> C, A, B;
	int t = 0;
	string g;
	for (int r = a.size() - 1; r >= 0; r--) A.push_back(a[r] - '0');
	for (int r = b.size() - 1; r >= 0; r--) B.push_back(b[r] - '0');
	for (int i = 0; i < A.size() || i < B.size(); i++)
	{
		if (i < A.size()) t += A[i];
		if (i < B.size()) t += B[i];
		C.push_back(t % 10);
		t /= 10;
	}
	if (t) C.push_back(1);
	for (int r = C.size() - 1; r >= 0; r--) g.push_back(C[r]+'0');
	return g;
}

int main()
{
	int n, x;
	cin >> n;
	string a, b, s;
	string str[210];
	vector<int> A, B;
	str[1] = "1";
	str[2] = "2";
	for (int i = 3; i < 210; i++)
		str[i] = add(str[i-1], str[i-2]);
	while (n--)
	{
		cin >> s;
		cout << str[s.size()] << endl;
	}
	
}