Fibonacci String

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4568    Accepted Submission(s): 1540

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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其实也完全可以用模拟来解决，不过非常麻烦2

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int dp[55][30];
char str1[35],str2[35];
int main(){
   int t;
   scanf("%d",&t);
   while(t--){
        memset(dp,0,sizeof(dp));
        memset(str1,0,sizeof(str1));
        memset(str2,0,sizeof(str2));
      getchar();
      int k;
      scanf("%s %s %d",str1,str2,&k);
      int len1=strlen(str1);
      int len2=strlen(str2);
      for(int i=0;i<len1;i++)
        dp[0][str1[i]-‘a‘]++;
      for(int i=0;i<len2;i++)
        dp[1][str2[i]-‘a‘]++;
      for(int i=2;i<=k;i++){
          for(int j=0;j<26;j++)
            dp[i][j]=dp[i-1][j]+dp[i-2][j];
      }
    for(int i=0;i<26;i++)
        printf("%c:%d\n",i+97,dp[k][i]);
    printf("\n");

   }
   return 0;
}