HDOJ(HDU) 1708 Fibonacci String

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

格式：每个案例后面都有一个空行！！！

不能直接用字符串相加来做，因为可能到后面会超内存！累加到后面的字符串太长了！！！

所以换位思考，既然是统计字母的个数，为什么不直接来建立整型数组呢。
只要统计出str0和str1中各个字母的个数就可以了。
后面各个字母个数的按照公式来推就行。

import java.util.Scanner;

public class Main{ public static void main(String[] args) { long num[][] = new long[56][26]; Scanner sc = new Scanner(System.in); int t = sc.nextInt(); while(t-->0){ for(int i=0;i<num[0].length;i++){ for(int j=0;j<num.length;j++){ num[j][i]=0; } } String str0 = sc.next(); for(int i=0;i<str0.length();i++){ for(int j='a';j<='z';j++){ if(str0.charAt(i)==(char)j){ num[0][j-'a']++; break; } } } String str1 = sc.next(); for(int i=0;i<str1.length();i++){ for(int j='a';j<='z';j++){ if(str1.charAt(i)==(char)j){ num[1][j-'a']++; break; } } } int n = sc.nextInt(); for(int i=2;i<=n;i++){ for(int k='a';k<='z';k++){ num[i][k-'a'] = num[i-1][k-'a']+num[i-2][k-'a']; } } for(int k='a';k<='z';k++){ System.out.println((char)k+":"+num[n][k-'a']); } System.out.println(); } }


}

  

 

  
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文章来源: chenhx.blog.csdn.net，作者：谙忆，版权归原作者所有，如需转载，请联系作者。

原文链接：chenhx.blog.csdn.net/article/details/51207917