Fibonacci String
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4825 Accepted Submission(s): 1626
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
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
#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int dp[3][26]={0};
void print11(int x);
int main()
{
char s1[35],s2[35];
int ka=0;
int t=0;
scanf("%d",&t);
getchar();
while(t--)
{
scanf("%s%s%d",s1,s2,&ka);
//scanf("%d",&ka);
getchar();
memset(dp,0,sizeof(dp));
//printf("1a\n");
int i=0;
while(s1[i++]) dp[0][int(s1[i-1]-'a')]++;
//printf("1a\n");
int j=0;
while(s2[j++]) dp[1][int(s2[j-1]-'a')]++;
//printf("1a\n");
for(i=2;i<=ka;i++)
{
for(j=0;j<26;j++)
dp[i%3][j]=dp[(i-1)%3][j]+dp[(i-2)%3][j];
}
print11(ka%3);
printf("\n");
}
return 0;
}
void print11(int x)
{
for(int i=0;i<26;i++)
{
printf("%c:%d\n",int('a'+i),dp[x][i]);
}
}