Fibonacci String

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

After little Jim learned Fibonacci Number in the class , he was very interest in it.

Now he is thinking about a new thing -- Fibonacci String .

He defines : str[n] = str[n-1] + str[n-2] ( n > 1 )

He is so crazying that if someone gives him two strings str[0] and str[1], he will calculate the str[2],str[3],str[4] , str[5]....

For example :

If str[0] = "ab"; str[1] = "bc";

he will get the result , str[2]="abbc", str[3]="bcabbc" , str[4]="abbcbcabbc" …………;

As the string is too long ,Jim can‘t write down all the strings in paper. So he just want to know how many times each letter appears in Kth Fibonacci String . Can you help him ?

Input

The first line contains a integer N which indicates the number of test cases.

Then N cases follow.

In each case,there are two strings str[0], str[1] and a integer K (0 <= K < 50) which are separated by a blank.

The string in the input will only contains less than 30 low-case letters.

Output

For each case,you should count how many times each letter appears in the Kth Fibonacci String and print out them in the format "X:N".

If you still have some questions, look the sample output carefully.

Please output a blank line after each test case.

To make the problem easier, you can assume the result will in the range of int.

Sample Input

1
ab bc 3

Sample Output

a:1
b:3
c:2
d:0
e:0
f:0
g:0
h:0
i:0
j:0
k:0
l:0
m:0
n:0
o:0
p:0
q:0
r:0
s:0
t:0
u:0
v:0
w:0
x:0
y:0
z:0

Author

linle

Source

HDU 2007-Spring Programming Contest

题解：求第你n个斐波那契字符串的字母统计.....求出第n个斐波那契字符串再统计。

AC代码：

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<iomanip>
#include<algorithm>
#include<time.h>
typedef long long LL;
using namespace std;

int main()
{
	int t;
    string s0,s1;
	int n;
	LL s[50]={0};
	s[1]=1;
	for(int i=2;i<50;i++)
	{
		s[i]=s[i-1]+s[i-2];
	}
	cin>>t;
	while(t--)
	{
		cin>>s0>>s1>>n;
		LL ch1[26]={0};
		LL ch2[26]={0};

		for(int i=0;i<s0.size();i++)
		{
			ch1[s0[i]-'a']++;    //字符转换成数字
		}
		for(int i=0;i<s1.size();i++)
		{
			ch2[s1[i]-'a']++;   //字符转换成数字
		}

		if(n==0)
		{
			for(int i=0;i<26;i++)
			{
				cout<<(char)('a'+i)<<":"<<ch1[i]<<endl;   //n=0时，只输出第一个
			}
			cout<<endl; continue;
		}
		for(int i=0;i<26;i++)
		{
			cout<<(char)('a'+i)<<":"<<s[n-1]*ch1[i]+s[n]*ch2[i]<<endl; //累加到第n-1个和第n个
		}
		cout<<endl;
	}

    return 0;

}