HDU 1708 Fibonacci String（数学题）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1708

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

PS:

记录斐波那契的数组当时傻逼了也开成了26，导致WA了几把都找不到原因！

图样图森破！

代码如下：

#include <cstdio>
#include <cstring>
int main()
{
    int t;
    char s0[30], s1[30];
    int k, a0[26], a1[26];
    int  c[50];
    c[0] = 0, c[1] = 1;
    for(int i = 2; i <= 50; i++)
    {
        c[i] = c[i-1] + c[i-2];
    }
    scanf("%d",&t);
    while(t--)
    {
        memset(s0,0,sizeof(s0));
        memset(s1,0,sizeof(s1));
        memset(a0,0,sizeof(a0));
        memset(a1,0,sizeof(a1));
        scanf("%s%s%d",s0,s1,&k);
        for(int i = 0; i < strlen(s0); i++)
        {
            int tt = s0[i]-'a';
            a0[tt]++;
        }
        for(int i = 0; i < strlen(s1); i++)
        {
            int tt = s1[i]-'a';
            a1[tt]++;
        }
        for(int i = 0; i < 26; i++)
        {
            if(k == 0)
            {
                printf("%c:%d\n",'a'+i,a0[i]);
            }
            else
                printf("%c:%d\n",'a'+i,a0[i]*c[k-1]+a1[i]*c[k]);
        }
        printf("\n");
    }
    return 0;
}