HDU 1708 Fibonacci String

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

 

Fibonacci String

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1556    Accepted Submission(s): 554

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

 

Recommend

lcy

 

分析：这个题，我刚刚看还真觉得麻烦，因为有26个字母，难道要int26个？最后看了网上的解题报告才知道，原来字符就是ASCII码，通过ASCII码用类似哈希打表的方式保存即可，再加上斐波那契数列递推就可以了。

代码如下：

#include<stdio.h>
#include<string.h>
int ans[50][27];                          //开空间要足够大，在这地方wrong了次.50 27应该是允许的最小值了
int main()
{
  int T,n,i,j;
  char s1[31],s2[31];
  scanf("%d",&T);
  while(T--)
  {
    scanf("%s%s%d",s1,s2,&n);
    memset(ans,0,sizeof(ans));
    for(i=0;s1[i]!=NULL;i++)
        ans[0][s1[i]-'a']++;
              for(i=0;s2[i]!=NULL;i++)
        ans[1][s2[i]-'a']++;
              for(i=2;i<=n;i++)
        for(j=0;j<26;j++)
                        ans[i][j]=ans[i-1][j]+ans[i-2][j];
              for(i=0;i<26;i++)
                    printf("%c:%d\n",'a'+i,ans[n][i]);
              printf("\n"); //Please output a blank line after each test case.又wrong了次。。。
  }
  return 0;
}