Fibonacci String HDU - 1708

Fibonacci String

  HDU - 1708 

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 ? 

InputThe 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. 
OutputFor 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

题解：

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
long long sum[55][30] = {0};
char s[2][10000];
int a[27] = {0};
int main()
{
	int t;
	scanf("%d", &t);
	while(t--)
	{
		memset(sum, 0, sizeof(sum));
		int n;
		scanf("%s %s %d", s[0], s[1], &n);
		for (int i = 0; i < 2; i++)
		{
			int l = strlen(s[i]);
			for (int j = 0; j < l; j++)
			{
				sum[i][s[i][j]-'a']++;
			}
		}
		for (int i = 2; i <= n; i++)
		{
			for(int j = 0; j < 26; j++)
			{
				sum[i][j] = sum[i-1][j] + sum[i-2][j];
			}
		}
		char c = 'a';
		for (int i = 0; i < 26; i++)
		{
			printf("%c:%lld\n", c+i, sum[n][i]);
		}
		printf("\n");
	}
	return 0;
}