hdu 1708 Fibonacci String

题目链接：Fibonacci String 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

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

看到题目我想都没想，直接上模拟，交了好几发都是mle，因为串太长了，然后看题解，发现大家基本都是用dp写的....dp[i][j]表示第i个数的j的值

参考http://blog.sina.com.cn/s/blog_96e2982001014tak.html

代码：

#include<stdio.h>#include<string.h>const int maxn=55;int dp[maxn][maxn];int main(){    //freopen("in.txt","r",stdin);    char a[50],b[50];    int k;    int t;    scanf("%d",&t);    while(t--)    {        memset(dp,0,sizeof(dp));        scanf("%s %s %d",a,b,&k);        for(int i=0;a[i];i++)            dp[0][a[i]-'a']++;        for(int i=0;b[i];i++)            dp[1][b[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+'a',dp[k][i]);        printf("\n");    }    return 0;}