1708 Fibonacci String

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

 

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <math.h>
using namespace std;
long long f(int n)//斐波那契数列函数
{
    long long i,a[100];
    a[0]=0;a[1]=1;
    for(i=2;i<=n;i++)
    a[i]=a[i-1]+a[i-2];
    return a[n];
}
int main()
{
     char str1[100],str2[100],ch;
     long long  n,m,i,j,k,sum,len1,len2;
     while(~scanf("%lld",&m))
     {
         while(n--)
         {
            scanf("%s%s%lld",str1,str2,&m);
            len1=strlen(str1);
            len2=strlen(str2);
            if(m==0)//一个特例
            {
                for(i=0;i<26;i++)
            {
                sum=0;ch='a'+i;
                for(j=0;j<len1;j++)
                {
                    if(ch==str1[j])
                    sum=sum+1;
                }
                printf("%c:%lld\n",ch,sum);
            }
            }
            else
            {
            for(i=0;i<26;i++)
            {
                sum=0;ch='a'+i;//找出从a~z
                for(j=0;j<len1;j++)
                {
                    if(ch==str1[j])//进行判断
                    sum=sum+f(m-1);//求出总和
                }
                for(j=0;j<len2;j++)
                {
                    if(ch==str2[j])
                    sum=sum+f(m);
                }
                printf("%c:%lld\n",ch,sum);//分别打印
            }
            }
           printf("\n");
         }
     }
     return 0;
}

这个问题中一个特点就是使用scanf输入，如果是cin的话就会超时。

另附一个从网上找到的代码：

#include<stdio.h>  
#include<string.h>  
char c[1000],s[1000];  
int a[27][100];//储存第1~100次所求字符串里边的第1~26个字母的个数.   
int main()  
{  
    int t,m,n,k,i,j;  
    scanf("%d",&t);  
    while(t--)  
    {  
        scanf("%s%s%d",c,s,&n);  
        int len=strlen(c);//测长度   
        int lem=strlen(s);  
        memset(a,0,sizeof(a));//清零a数组.   
        for(j=0;j<len;j++)  
        for(i=1;i<=26;i++)  
        if(c[j]==i+'a'-1)//如果当前字符等于第i个字母   
        a[i][1]++;//则在a[i][1]++；   
        for(j=0;j<lem;j++)  
        for(i=1;i<=26;i++)  
        if(s[j]==i+'a'-1)  
        a[i][2]++;  //同理得到第二个字符串的 每一个字母有多少个.   
        for(i=1;i<=26;i++)  
        for(j=3;j<=n+1;j++)  
        a[i][j]=a[i][j-1]+a[i][j-2];//进行斐波那契相加.   
        for(i=1;i<=26;i++)  
        printf("%c:%d\n",i+'a'-1,a[i][n+1]);  
        printf("\n");//每一次样例后需要加一个换行，因为没看这个pe了一次.   
    }  
    return 0;  
}