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
= 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
[k-'a']);
}
System.out.println();

}
}

}