hdu 1708

Fibonacci String

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

Total Submission(s): 2079 Accepted Submission(s): 727



[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

[align=left]Author[/align]
linle

[align=left]Source[/align]
HDU 2007-Spring Programming Contest

[align=left]Recommend[/align]
lcy

//该题思路：用二维数组，一个用于递推，一个用于统计每个字母数。同时灵活运用字符ASCII码值。
#include <stdio.h>
#include <string.h>
int main()
{
int n, k, f[51][27], i, j;
char str1[31], str2[31];
scanf("%d", &n);
while (n--)
{
scanf("%s%s%d", str1, str2, &k);
memset(f, 0, sizeof(f));
for(i = 0; str1[i] != NULL; i++)
f[0][str1[i] - 'a']++;
for(i = 0; str2[i] != NULL; i++)
f[1][str2[i] - 'a']++;
for(i = 2; i <= k; i++)
for (j = 0; j < 26; j++)
{
f[i][j] = f[i-1][j] + f[i-2][j];
}
for(i = 0; i < 26; i++)
printf("%c:%d\n", i + 'a', f[k][i]);
printf("\n");
}
return 0;
}