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hdu 1708

2015-08-15 17:14 295 查看

http://acm.hdu.edu.cn/showproblem.php?pid=1708

Fibonacci String

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 4653 Accepted Submission(s): 1568

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

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
char  str[2][100];
int main()
{
    long long ans[50][30];
    int t,k;
    cin>>t;
    while(t--)
    {
          scanf("%s%s%d",str[0],str[1],&k);//  cin>>str[0]>>str[1]>>k;
        memset(ans,0,sizeof(ans));
        for(int i=0; i<strlen(str[0]); i++)
             ans[0][str[0][i]-'a']++;
        for(int i=0; i<strlen(str[1]); i++)
            ans[1][str[1][i]-'a']++;
        for(int i=2; i<50; i++)
            for(int j=0; j<26; j++)
                ans[i][j]=ans[i-1][j]+ans[i-2][j];
        char c='a';
        for(int i=0; i<26; i++)
        printf("%c:%lld\n",c+i,ans[k][i]);
        cout<<endl;
    }
    return 0;
}

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