Fibonacci String(hdu 1708)

Fibonacci String

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5008    Accepted Submission(s): 1690


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
 

 

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水题，简单的递推。
 1 #include<iostream>
 2 #include<string.h>
 3 #include<queue>
 4 #include<stdio.h>
 5 #include<algorithm>
 6 using namespace std;
 7 char a[40];
 8 char b[40];
 9 int aa[26];
10 int bb[26];
11 int cc[26];
12 int main(void)
13 {
14     int n,i,j,k,p,q;
15     scanf("%d",&k);
16     while(k--)
17     {
18         scanf("%s %s",a,b);
19         memset(aa,0,sizeof(aa));
20         memset(bb,0,sizeof(bb));
21         scanf("%d",&p);
22         int l=strlen(a);
23         int r=strlen(b);
24         for(i=0;i<l;i++)
25         {aa[a[i]-'a']+=1;
26         }
27 
28         for(i=0;i<r;i++)
29         {
30             bb[b[i]-'a']+=1;
31         }
32         if(p==0)
33         {
34             for(i=0;i<=25;i++)
35             {
36                 printf("%c:",i+'a');
37                 printf("%d\n",aa[i]);
38             }
39         }
40         else if(p==1)
41         {
42                         for(i=0;i<=25;i++)
43             {
44                 printf("%c:",i+'a');
45                 printf("%d\n",bb[i]);
46             }
47         }
48         else
49         {
50             for(i=0;i<p-1;i++)
51             {
52                 for(j=0;j<=25;j++)
53                 {
54                     cc[j]=aa[j]+bb[j];
55                 }
56                 for(j=0;j<=25;j++)
57                 {
58                     aa[j]=bb[j];
59                 }
60                 for(j=0;j<=25;j++)
61                 {
62                     bb[j]=cc[j];
63                 }
64             }
65             for(i=0;i<=25;i++)
66             {
67               printf("%c:",i+'a');
68               printf("%d\n",cc[i]);
69             }
70         }printf("\n");
71 
72     }
73     return 0;
74 }

 

posted @ 2016-01-23 12:36  sCjTyC  阅读(316)  评论(0)  编辑  收藏  举报