Fibonacci String

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5012 Accepted Submission(s):
1693

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
 
Recommend
lcy | We have carefully selected several similar
problems for you: 1709 1710 1707 1701 1714
 
很简单的模拟题,注意给的第一个数是s0
 
题意:给两个字符串s0和s1,sn是sn-1和sn-2拼接而成的。问sn中每个字母出现的次数
 
附上代码:
 
 #include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int main()
{
int f[][];
char s1[],s2[];
int i,j,T,n,m;
scanf("%d",&T);
while(T--)
{
scanf("%s %s %d",s1,s2,&n);
memset(f,,sizeof(f));
for(i=; s1[i]!='\0'; i++)
f[][s1[i]-'a']++; //字母转化为数字保存
for(i=; s2[i]!='\0'; i++)
f[][s2[i]-'a']++;
for(i=; i<=n; i++)
{
for(j=; j<; j++)
f[i][j]=f[i-][j]+f[i-][j]; //每一次都是前面两个相加
}
for(i=; i<; i++)
printf("%c:%d\n",i+'a',f[n][i]); //输出26个字母的个数
printf("\n");
}
return ;
}

最新文章

  1. java hook
  2. Ubuntu 14.10安装SecureCRT 7.3(转)
  3. 在VisualStudio中应该使用什么字体
  4. hdu 5058 So easy
  5. 禁用nginx的access日志
  6. TF卡的Class4/Class6/Class10
  7. RBF径向基神经网络——乳腺癌医学诊断建模
  8. 转: seajs手册与文档之--模块定义
  9. 基于visual Studio2013解决面试题之0306打印第一次只出现一次的字符
  10. WPF触屏Touch事件在嵌套控件中的响应问题
  11. javascript闭包小结
  12. FastJson将json解析成含有泛型对象,内部泛型对象再次解析出错的解决办法(Android)
  13. 原型模式(Prototype)
  14. Apache配置虚拟主机后,不能访问localhost
  15. 使用Jmeter自带的 Http 代理服务器录制脚本
  16. python—函数
  17. 《大话设计模式》c++实现 模版方法模式
  18. 20145221高其_Web基础
  19. 写一个小程序实现win系统定时锁屏
  20. JQuery效果隐藏/显示

热门文章

  1. Katalon系列二十一:用例中语句失败处理
  2. React map生成元素添加点击事件绑定this
  3. Redis 扫盲
  4. EMAS,一部淘宝十年移动互联网技术的演进史
  5. Servlet容器container
  6. 【JZOJ4934】【NOIP2017GDKOI模拟1.12】a
  7. 【JZOJ4815】【NOIP2016提高A组五校联考4】ksum
  8. ORACLE 日常维护命令手册
  9. springboot 数据访问【转】【补】
  10. 使用 docker-compose 安装 MySQL 5.5 记录