Fibonacci String 1708

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

1ab bc 3

Sample Output

a:1b:3c:2d:0e:0f:0g:0h:0i:0j:0k:0l:0m:0n:0o:0p:0q:0r:0s:0t:0u:0v:0w:0x:0y:0z:0
AC代码
#include <cstdio>#include <map>#include <string>#include <algorithm>#include <iostream>int main(int argc, const char* argv[]){    int nCases = 0;    scanf("%d", &nCases);    while (nCases--)    {        int k;        std::string sz0, sz1;        std::cin >> sz0 >> sz1 >> k;        ///初始化第一个与第二个        std::map<char, __int64> map_a, map_b, map_c;        for (char ch= 'a'; ch <= 'z'; ++ch)        {            map_a[ch] = count(sz0.begin(), sz0.end(), ch);            map_b[ch] = count(sz1.begin(), sz1.end(), ch);        }        ///模拟向后加        for (int i=2; i<=k; ++i)        {            if (i%3 == 2)            {                for (char ch= 'a'; ch <= 'z'; ++ch)                {                    map_c[ch] = map_a[ch] + map_b[ch];                }            }            if (i%3 == 0)            {                for (char ch= 'a'; ch <= 'z'; ++ch)                {                    map_a[ch] = map_b[ch] + map_c[ch];                }            }            if (i%3 == 1)            {                for (char ch= 'a'; ch <= 'z'; ++ch)                {                    map_b[ch] = map_a[ch] + map_c[ch];                }            }        }        ///输出结果        if (k%3 == 0)        {            for (char ch= 'a'; ch <= 'z'; ++ch)            {                printf("%c:%I64d\n", ch, map_a[ch]);            }        }        else if (k%3 == 1)        {            for (char ch= 'a'; ch <= 'z'; ++ch)            {                printf("%c:%I64d\n", ch, map_b[ch]);            }        }        else        {            for (char ch= 'a'; ch <= 'z'; ++ch)            {                printf("%c:%I64d\n", ch, map_c[ch]);            }        }        printf("\n");    }    return 0;}