Light oj 1105 - Fi Binary Number（计数）

本文链接：https://blog.youkuaiyun.com/u014737310/article/details/48661509

本文介绍了一种特殊的二进制数——Fi-Binary数，并提供了一个算法来计算第n个Fi-Binary数。Fi-Binary数的特点是不含前导0且不包含连续的1。文章通过示例解释了输入输出格式。

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1105 - Fi Binary Number

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Time Limit: 2 second(s)

Memory Limit: 32 MB

A Fi-binary number is a number that contains only 0 and 1. It does not contain any leading 0. And also it does not contain 2 consecutive 1. The first few such number are 1, 10, 100, 101, 1000, 1001, 1010, 10000, 10001, 10010, 10100, 10101 and so on. You are given n. You have to calculate the n^th Fi-Binary number.

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case contains an integer n (1 ≤ n ≤ 10⁹).

Output

For each case, print the case number and the n^th Fi-Binary number

Sample Input

Output for Sample Input

Case 1: 10010

Case 2: 101010

Case 3: 1010001

Case 4: 10001001

PROBLEM SETTER: ABDULLAH AL MAHMUD

SPECIAL THANKS: JANE ALAM JAN (SOLUTION, DATASET)

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define bug printf("hihi\n")

#define eps 1e-12

typedef long long ll;
using namespace std;

#define mod 1000000007
#define N 100

ll f[2][N];

void inint()
{
    int i,j;
    int one,zero;
    zero=1;
    one=0;
    f[0][0]=f[1][0]=0;
    for(i=1;i<N;i++)
       {
           f[0][i]=f[0][i-1]+one+zero;
           int te=one;
           one=zero;
           zero=zero+te;
       }
    zero=0;
    one=1;
    for(i=1;i<N;i++)
    {
        f[1][i]=f[1][i-1]+one+zero;
        int te=one;
        one=zero;
        zero=zero+te;
    }
}

void dfs(int st,int n)
{

    if(n==0) return ;

    if(st==1)
    {
        printf("1");
        dfs(0,n-1);
        return ;
    }

    printf("0");
    n--;
    if(n==0) return ;

    int i;
    for(i=1;i<N;i++)
        if(f[0][i]+f[1][i]>=n) break;

    if(n<=f[0][i]+f[1][i-1])
    {
        dfs(0,n-f[1][i-1]);
        return ;
    }
    dfs(1,n-f[0][i]);
}

int main()
{
    int i,j,t,ca=0,n;
    inint();
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        printf("Case %d: ",++ca);
        dfs(1,n);
        printf("\n");
    }
    return 0;
}