hduoj 1708

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2722    Accepted Submission(s): 920

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

 

#include<stdio.h>
#include<string.h>
int ans[50][27];//由于有26个字母               
int main()
{
 int T,n,i,j;
 char s1[31],s2[31];
 scanf("%d",&T);
 while(T--)
 {
  scanf("%s%s%d",s1,s2,&n);
  memset(ans,0,sizeof(ans));//将数组全为0；
  for(i=0;s1[i]!=NULL;i++)
     ans[0][s1[i]-'a']++;//统计字母数
        for(i=0;s2[i]!=NULL;i++)
     ans[1][s2[i]-'a']++;//统计字母数
        for(i=2;i<=n;i++)
     for(j=0;j<26;j++)
             ans[i][j]=ans[i-1][j]+ans[i-2][j];// n次 字母次数相加
        for(i=0;i<26;i++)
           printf("%c:%d\n",'a'+i,ans[n][i]);
        printf("\n"); 
 }
 return 0;
}