HDU 1708 简单dp问题 Fibonacci String

Fibonacci String

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4568    Accepted Submission(s): 1540


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
 

 

Author
linle
 

 

Source
 

 

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其实也完全可以用模拟来解决,不过非常麻烦2
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int dp[55][30];
char str1[35],str2[35];
int main(){
   int t;
   scanf("%d",&t);
   while(t--){
        memset(dp,0,sizeof(dp));
        memset(str1,0,sizeof(str1));
        memset(str2,0,sizeof(str2));
      getchar();
      int k;
      scanf("%s %s %d",str1,str2,&k);
      int len1=strlen(str1);
      int len2=strlen(str2);
      for(int i=0;i<len1;i++)
        dp[0][str1[i]-'a']++;
      for(int i=0;i<len2;i++)
        dp[1][str2[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+97,dp[k][i]);
    printf("\n");
    
   }
   return 0;
}

 

posted @ 2015-07-22 19:14  柳下_MBX  阅读(416)  评论(0编辑  收藏  举报