Revenge of Fibonacci
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 204800/204800 K (Java/Others)
Total Submission(s): 2405 Accepted Submission(s): 609
Problem Description
The well-known Fibonacci sequence is defined as following:
Here we regard n as the index of the Fibonacci number F(n).
This sequence has been studied since the publication of Fibonacci‘s book Liber Abaci. So far, many properties of this sequence have been introduced.
You had been interested in this sequence, while after reading lots of papers about it. You think there’s no need to research in it anymore because of the lack of its unrevealed properties. Yesterday, you decided to study some other sequences like Lucas sequence
instead.
Fibonacci came into your dream last night. “Stupid human beings. Lots of important properties of Fibonacci sequence have not been studied by anyone, for example, from the Fibonacci number 347746739…”
You woke up and couldn’t remember the whole number except the first few digits Fibonacci told you. You decided to write a program to find this number out in order to continue your research on Fibonacci sequence.
Input
There are multiple test cases. The first line of input contains a single integer T denoting the number of test cases (T<=50000).
For each test case, there is a single line containing one non-empty string made up of at most 40 digits. And there won’t be any unnecessary leading zeroes.
Output
For each test case, output the smallest index of the smallest Fibonacci number whose decimal notation begins with the given digits. If no Fibonacci number with index smaller than 100000 satisfy that condition, output -1 instead – you think what Fibonacci
wants to told you beyonds your ability.
Sample Input
15 1 12 123 1234 12345 9 98 987 9876 98765 89 32 51075176167176176176 347746739 5610
Sample Output
Case #1: 0 Case #2: 25 Case #3: 226 Case #4: 1628 Case #5: 49516 Case #6: 15 Case #7: 15 Case #8: 15 Case #9: 43764 Case #10: 49750 Case #11: 10 Case #12: 51 Case #13: -1 Case #14: 1233 Case #15: 22374
Source
2011 Asia Shanghai Regional Contest
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ac代码
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
char a[100],b[100],c[100],t[100];
typedef struct s
{
struct s * child[10];
int id;
}node,*Node;
Node root;
void add(char a[],char b[],char c[])
{
int len1,len2,i,j,t[10000],max,k=0;
len1=strlen(a);
len2=strlen(b);
i=len1-1;
j=len2-1;
memset(t,0,sizeof(t));
while(i>=0||j>=0)
{
if(i<0&&j>=0)
t[k]+=b[j]-'0';
else
if(j<0&&i>=0)
t[k]+=a[i]-'0';
else
{
t[k]+=a[i]-'0'+b[j]-'0';
}
k++;
t[k]+=t[k-1]/10;
t[k-1]%=10;
if(t[k])
max=k;
else
max=k-1;
i--;
j--;
}
for(i=max;i>=0;i--)
c[max-i]=t[i]+'0';
c[max+1]='\0';
//return c;
}
char str[3][100];
void insert(char *s,int id)
{
Node cur,newnode;
int i,now,len;
len=strlen(s);
cur=root;
//int i;
for(i=0;i<len&&i<41;i++)
{
now=s[i]-'0';
if(cur->child[now]!=NULL)
{
cur=cur->child[now];
}
else
{
newnode=(Node)calloc(1,sizeof(node));
cur->child[now]=newnode;
cur=cur->child[now];
cur->id=id;
}
}
}
int seach(char *s)
{
int len=strlen(s),i;
Node cur;
cur=root;
for(i=0;i<len;i++)
{
int now=s[i]-'0';
if(cur->child[now]==NULL)
break;
cur=cur->child[now];
}
if(i<len)
return -1;
return cur->id;
}
void fun()
{
strcpy(a,"1");
insert(a,0);
strcpy(b,"1");
insert(b,1);
for(int i=2;i<100000;i++)
{
int len1=strlen(a);
int len2=strlen(b);
if(len2>60)
{
b[len2-1]=0;
a[len1-1]=0;
}
add(a,b,t);
insert(t,i);
strcpy(a,b);
strcpy(b,t);
// if(i<30)
// printf("%s\n",t);
}
}
int main()
{
int t,c=0;
root=(Node)calloc(1,sizeof(node));
fun();
scanf("%d",&t);
while(t--)
{
char temp[50];
scanf("%s",temp);
printf("Case #%d: ",++c);
printf("%d\n",seach(temp));
}
}
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