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

1ab bc 3

 

Sample Output

a:1b:3c:2d:0e:0f:0g:0h:0i:0j:0k:0l:0m:0n:0o:0p:0q:0r:0s:0t:0u:0v:0w:0x:0y:0z: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;}