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 }