hdu1708——Fibonacci String

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

斐波那其字符串,求第k个字符串中每个字母出现的个数,用一个二维数组解决

代码:

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN=50;
int dp[MAXN][MAXN];
int main(void){
    int t;
    char a[55];
    char b[55];
    scanf("%d",&t);
    while(t--){
        int k;
        scanf("%s %s %d",a,b,&k);
        memset(dp,0,sizeof(dp));
        for(int i=0;a[i]!='\0';i++){
            dp[0][a[i]-'a']++;
        }
        for(int i=0;b[i]!='\0';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;
}

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转载自blog.csdn.net/westbrook1998/article/details/80349124
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