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 .
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.
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.
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
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
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
Recommend
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[][];
char str1[],str2[];
int main(){
int t;
scanf("%d",&t);
while(t--){
memset(dp,,sizeof(dp));
memset(str1,,sizeof(str1));
memset(str2,,sizeof(str2));
getchar();
int k;
scanf("%s %s %d",str1,str2,&k);
int len1=strlen(str1);
int len2=strlen(str2);
for(int i=;i<len1;i++)
dp[][str1[i]-'a']++;
for(int i=;i<len2;i++)
dp[][str2[i]-'a']++;
for(int i=;i<=k;i++){
for(int j=;j<;j++)
dp[i][j]=dp[i-][j]+dp[i-][j];
}
for(int i=;i<;i++)
printf("%c:%d\n",i+,dp[k][i]);
printf("\n"); }
return ;
}