hdu 1708 Fibonacci String

Fibonacci String

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5012 Accepted Submission(s):
1693

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 | We have carefully selected several similar
problems for you: 1709 1710 1707 1701 1714

 

很简单的模拟题，注意给的第一个数是s0

 

题意：给两个字符串s0和s1，sn是sn-1和sn-2拼接而成的。问sn中每个字母出现的次数

 

附上代码：

 

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 using namespace std;
 int main()
 {
     int f[][];
     char s1[],s2[];
     int i,j,T,n,m;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%s %s %d",s1,s2,&n);
         memset(f,,sizeof(f));
         for(i=; s1[i]!='\0'; i++)
             f[][s1[i]-'a']++;            //字母转化为数字保存
         for(i=; s2[i]!='\0'; i++)
             f[][s2[i]-'a']++;
         for(i=; i<=n; i++)
         {
             for(j=; j<; j++)
                 f[i][j]=f[i-][j]+f[i-][j]; //每一次都是前面两个相加
         }
         for(i=; i<; i++)
             printf("%c:%d\n",i+'a',f[n][i]);  //输出26个字母的个数
         printf("\n");
     }
     return ;
 }