1708 Fibonacci String

[align=left]Problem Description[/align]
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
= 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 ?

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

1
ab bc 3

[align=left]Sample Output[/align]

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

#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;
}

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