HDU 1708 hash

Fibonacci String

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

Total Submission(s): 4825    Accepted Submission(s): 1626


[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<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int dp[3][26]={0};
void print11(int x);
int main()
{
 char s1[35],s2[35];
 int ka=0;
 int t=0;
 scanf("%d",&t);
 getchar();
 while(t--)
 {
  scanf("%s%s%d",s1,s2,&ka);
  //scanf("%d",&ka);
  getchar();
  memset(dp,0,sizeof(dp));
  //printf("1a\n");
  int i=0;
  while(s1[i++]) dp[0][int(s1[i-1]-'a')]++;
  //printf("1a\n");
  int j=0;
  while(s2[j++]) dp[1][int(s2[j-1]-'a')]++;
  //printf("1a\n");
   for(i=2;i<=ka;i++)
   {
      for(j=0;j<26;j++)
    dp[i%3][j]=dp[(i-1)%3][j]+dp[(i-2)%3][j];
   }
   print11(ka%3);
   printf("\n");
 }
 return 0;
}
void print11(int x)
{
 for(int i=0;i<26;i++)
 {
  printf("%c:%d\n",int('a'+i),dp[x][i]);
 }
}