UVA11089 Fi-binary Number数学规律

Posted 2021-02-09 tigerisland45

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
The first line of the input contains one integer T the number of test cases. Each test case contains one integer n.
Output
For each test case output one line containing the n-th Fi-Binary number.
Constraints
? 1 ≤ N ≤ 10^9
Sample Input
4
10
20
30
40
Sample Output
10010
101010
1010001
10001001

问题链接：UVA11089 Fi-binary Number
问题简述：（略）
问题分析：
????Fi-binary Number数定义为由0和1构成，开头不是0，且没有连续的两个1。该数的数列依次是1, 10, 100, 101, 1000, 1001,1010, 10000, 10001, 10010, 10100, 10101, ......。对于给定的n，计算该数列的第n项。
????该数列的长度个数正好呈现菲波那契数列特性，利用该特性进行计算。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* UVA11089 Fi-binary Number */

#include <iostream>

using namespace std;

const int N = 43;
int f[N + 1];

void setfib()
{
    f[0] = 1;
    f[1] = 1;
    for(int i = 2; i <= N; i++)
            f[i] = f[i - 2] + f[i - 1];
}

void solve(int n)
{
    n--;
    int k;
    for (k = 0; f[k] <= n; k++)
        n -= f[k];

    printf("1");
    while (k > 1) {
        if (n < f[k - 1]) {
            printf("0");
            k--;
        } else {
            printf("01");
            n -= f[k - 1];
            k -= 2;
        }
    }

    printf("%s
", k ? "0" : "");
}

int main()
{
    setfib();

    int t, n;
    scanf("%d", &t);
    while(t--) {
        scanf("%d", &n);

        solve(n);
    }

    return 0;
}