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：

 

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5 int main()
 6 {
 7     int f[50][30];
 8     char s1[35],s2[35];
 9     int i,j,T,n,m;
10     scanf("%d",&T);
11     while(T--)
12     {
13         scanf("%s %s %d",s1,s2,&n);
14         memset(f,0,sizeof(f));
15         for(i=0; s1[i]!='\0'; i++)
16             f[0][s1[i]-'a']++;            //字母转化为数字保存
17         for(i=0; s2[i]!='\0'; i++)
18             f[1][s2[i]-'a']++;
19         for(i=2; i<=n; i++)
20         {
21             for(j=0; j<26; j++)
22                 f[i][j]=f[i-1][j]+f[i-2][j]; //每一次都是前面两个相加
23         }
24         for(i=0; i<26; i++)
25             printf("%c:%d\n",i+'a',f[n][i]);  //输出26个字母的个数
26         printf("\n");
27     }
28     return 0;
29 }