Fibonacci String
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4102 Accepted Submission(s): 1396
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
HDU 2007-Spring Programming Contest
一开始就是挺傻的 ，最后才发现必然超内存。数据量是超级大！参考网上代码，用到了映射 ，顺便也学习了下。解题思路是用三个map组 分别记录相邻的三个状态。并对每一次字符串叠加更新每个字母的状态！
开始的时候 错误代码，头脑太简单了。：
#include<iostream>
#include<string>
using namespace std;
int main()
{
int n,k,ls[30];
string str[51];
cin>>n;
while(n--)
{
for(int l=0;l<30;l++)
ls[l]=0;
cin>>str[0]>>str[1]>>k;
for(int i=2;i<=k;i++)
str[i]=str[i-2]+str[i-1];
for(int m=0;m<str[k].size();m++)
{
ls[str[k][m]-'a']++;
}
for(int t=0;t<26;t++)
{
cout<<char('a'+t)<<":"<<ls[t]<<endl;
}
cout<<endl;
}
return 0;
}
AC Code：
#include<iostream>
#include<map>
#include<string>
using namespace std;
int main()
{
int n;
cin>>n;
while(n--)
{
map<char,int>ls1;
map<char,int>ls2;
map<char,int>ls3;
string str0,str1;
int k;
cin>>str0>>str1>>k;
if(k==0)
{
for(int i=0;i<str0.size();i++)
ls1[str0[i]]++;
for( int g='a';g<='z';g++)
cout<<char(g)<<":"<<ls1[g]<<endl;
}
else if(k==1)
{
for(int m=0;m<str1.size();m++)
ls2[str1[m]]++;
for(int k='a';k<='z';k++)
cout<<char(k)<<":"<<ls2[k]<<endl;
}
else
{
for(int q=0;q<str0.size();q++)
ls1[str0[q]]++;
for(int v=0;v<str1.size();v++)
ls2[str1[v]]++;
while(--k)
{
for(int p=97;p<=122;p++)
{
ls3[p]=ls2[p]+ls1[p];
ls1[p]=ls2[p];
ls2[p]=ls3[p];
}
}
for(int M=97;M<123;M++)
cout<<char(M)<<":"<<ls3[M]<<endl;
}
cout<<endl;
}
return 0;
}
又看了看，发现了一个写的比上个思路更加简洁的。来自：
“梦醒的博客”
用了个二维数组。
请看：
#include<stdio.h>
#include<string.h>
int ans[50][27];
int main()
{
int T,n,i,j;
char s1[31],s2[31];
scanf("%d",&T);
while(T--)
{
scanf("%s%s%d",s1,s2,&n);
memset(ans,0,sizeof(ans));
for(i=0;s1[i]!=NULL;i++)
ans[0][s1[i]-'a']++;
for(i=0;s2[i]!=NULL;i++)
ans[1][s2[i]-'a']++;
for(i=2;i<=n;i++)
for(j=0;j<26;j++)
ans[i][j]=ans[i-1][j]+ans[i-2][j];
for(i=0;i<26;i++)
printf("%c:%d\n",'a'+i,ans[n][i]);
printf("\n");
}
return 0;
}