首页 > ACM题库 > HDU-杭电 > HDU 1708 Fibonacci String-字符串-[解题报告] C++
2013
12-21

HDU 1708 Fibonacci String-字符串-[解题报告] C++

Fibonacci String

问题描述 :

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 ?

输入:

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.

输出:

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.

样例输入:

1
ab bc 3

样例输出:

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>
#include <cmath>
using namespace std;
long long tag[51][26];
char s1[31],s2[31];
int n;
void Init(char *s1,char *s2)
{
    memset(tag,0,sizeof(tag));
    for(int i=0;s1[i];i++)
        tag[0][s1[i]-'a']++;
    for(int i=0;s2[i];i++)
        tag[1][s2[i]-'a']++;
}
void solve()
{
    Init(s1,s2);
    for(int i=2;i<=n;i++)
    {
        for(int j=0;j<26;j++)
            tag[i][j]=tag[i-2][j]+tag[i-1][j];
    }
    for(int i=0;i<26;i++)
        printf("%c:%lld\n",'a'+i,tag[n][i]);
    printf("\n");
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s%s%d",s1,s2,&n);
        solve();
    }
    return 0;
}

解题报告转自:http://blog.csdn.net/shiyuankongbu/article/details/8457381


  1. 这道题目的核心一句话是:取还是不取。
    如果当前取,则index+1作为参数。如果当前不取,则任用index作为参数。

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