hdu 1708 Fibonacci String 大水題 2種大水法

Fibonacci String

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1969    Accepted Submission(s): 690

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  

題意： 不解釋 太水了

#include<stdio.h>
#include<string.h>
int main()
{
	int a[30],b[30];
	char s1[50],s2[50];
	int cas,n,k,i,flag,flag1;
	scanf("%d",&cas);
	while(cas--)
	{
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		scanf("%s %s %d",s1,s2,&k);
		if(k%2==0) flag=1;// 偶數輸出a[i] 奇數輸出b[i] 具體原因看下面
		else flag=0;
		for(i=0;i<strlen(s1);i++)
		{
			a[s1[i]-'a']++;
		}
		if(k==0)
		{
			for(i=0;i<26;i++)
				printf("%c:%d\n",i+'a',a[i]);
			printf("\n");continue;
		}
		for(i=0;i<strlen(s2);i++)
			b[s2[i]-'a']++;
		if(k==1)
		{
			for(i=0;i<26;i++)
				printf("%c:%d\n",i+'a',b[i]);
			printf("\n");continue;
		}
		k=k-1;
		flag1=1;
		while(k--)
		{
			if(flag1==1)
			{
		    	for(i=0;i<26;i++)
				{
				     a[i]=b[i]+a[i];
				}
				flag1=!flag1;
			}
			else
			{
			    for(i=0;i<26;i++)
				{
				    b[i]=b[i]+a[i];
				}
				flag1=!flag1;
			}

		}
		if(flag)
		{
			for(i=0;i<26;i++)
				printf("%c:%d\n",i+'a',a[i]);
		}
		else
		{
			for(i=0;i<26;i++)
				printf("%c:%d\n",i+'a',b[i]);
		}
		if(cas!=0)
		    printf("\n");

	}
	return 0;
}
           

自己的方法有點麻煩   看了下别人的代碼  哎呦 自己的那叫一個複雜啊

參考代碼作者

夢醒之後，燈火闌珊

#include<stdio.h>
#include<string.h>
int ans[50][27];             
int main()
{
 int T,n,i,j;
 char s1[31],s2[31];
 scanf("%d",&T);
 while(T--)
 {
  scanf("%s%s%d",s1,s2,&n);
  memset(ans,0,sizeof(ans));
  for(i=0;s1[i]!=NULL;i++)
     ans[0][s1[i]-'a']++;
        for(i=0;s2[i]!=NULL;i++)
     ans[1][s2[i]-'a']++;
        for(i=2;i<=n;i++)
     for(j=0;j<26;j++)
             ans[i][j]=ans[i-1][j]+ans[i-2][j];
        for(i=0;i<26;i++)
           printf("%c:%d\n",'a'+i,ans[n][i]);
        printf("\n"); 
 }
 return 0;
}