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

好坏好坏的一道题，不知道算不算水题，感觉这道题还挺好的，斐波那契字符串，就是说第0串字符为 ab，第1串为bc，那么根据斐波那契第二串字符为abbc，第三串为bcabbc因为第三串里边有1个a，三个b，两个c,别的没有了，所以输出了测试样例那样的.

要是按照斐波那契一个一个字符串的加，估计输入一个50，最后一个的字符串都不知道多长了，同时字符数组也开不了这么大，于是就开始分析，26个字母里边的每一个字母也是按斐波那契规律增长的，比如说，a在第0个字符串里边是1次，第1个字符串里边是0次，第二个里边是1+0=1次，第三次就是1+0=1次，同理得到了样例输出的b:3;

所以根据原理，就可以写出简单的代码.

附ac代码：

#include<stdio.h>
#include<string.h>
char c[1000],s[1000];
int a[27][100];//储存第1~100次所求字符串里边的第1~26个字母的个数.
int main()
{
	int t,m,n,k,i,j;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%s%s%d",c,s,&n);
		int len=strlen(c);//测长度
		int lem=strlen(s);
		memset(a,0,sizeof(a));//清零a数组.
		for(j=0;j<len;j++)
		for(i=1;i<=26;i++)
		if(c[j]==i+'a'-1)//如果当前字符等于第i个字母
		a[i][1]++;//则在a[i][1]++；
		for(j=0;j<lem;j++)
		for(i=1;i<=26;i++)
		if(s[j]==i+'a'-1)
		a[i][2]++;	//同理得到第二个字符串的 每一个字母有多少个.
		for(i=1;i<=26;i++)
		for(j=3;j<=n+1;j++)
		a[i][j]=a[i][j-1]+a[i][j-2];//进行斐波那契相加.
		for(i=1;i<=26;i++)
		printf("%c:%d\n",i+'a'-1,a[i][n+1]);
		printf("\n");//每一次样例后需要加一个换行，因为没看这个pe了一次.
	}
	return 0;
}

附测试样例：

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