学步园

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 ?

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.

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.

1
ab bc 3

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

题意很明白：思路见代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

int main()
{
    int t;
    char s1[50],s2[50],s;
    int a[100],b[100],c[100];
    int len1,len2,f;
    int n,i,j;
    scanf("%d",&t);
    while(t--)
    {
        memset(s1,0,sizeof(s1));
        memset(s2,0,sizeof(s2));
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        memset(c,0,sizeof(c));
        scanf("%s%s%d",s1,s2,&n);
        len1=strlen(s1);
        len2=strlen(s2);
        for(i=0; i<len1+len2; i++)
        {
            if(s1[i]!='\0')
            {
                a[s1[i]-97]++;
                // b[s2[i]-96]++;
            }
            if(s2[i]!='\0')
            {
                b[s2[i]-97]++;
            }
        }
        f=1;
        for(i=2; i<=n; i++)
        {
            for(j=0; j<26; j++)
            {
                c[j]=a[j]+b[j];
            }
            if(f==1)
            {
                for(j=0; j<26; j++)
                {
                    a[j]=c[j];
                }
                f=0;
            }
            else
            {
                for(j=0; j<26; j++)
                {
                    b[j]=c[j];
                }
                f=1;
            }
        }
        s='a';
        if(n==0)
        {
            for(i=0; i<26; i++)
            {
                printf("%c:%d\n",s++,a[i]);
            }
            printf("\n");
            continue;
        }
        if(n==1)
        {
            for(i=0; i<26; i++)
            {
                printf("%c:%d\n",s++,b[i]);
            }
            printf("\n");
            continue;
        }
        for(i=0; i<26; i++)
        {
            printf("%c:%d",s++,c[i]);
            printf("\n");
        }
        printf("\n");
    }
    return 0;
}