ACM学习历程——HDU4814 Golden Radio Base（数学递推）（12年成都区域赛）

时间：2014-10-23 20:28:51 阅读：350 评论：0 收藏：0 [点我收藏+]

Description

Golden ratio base (GRB) is a non-integer positional numeral system that uses the golden ratio (the irrational number (1+√5)/2 ≈ 1.61803399 symbolized by the Greek letter φ) as its base. It is sometimes referred to as base-φ, golden mean base, phi-base, or, phi-nary.
Any non-negative real number can be represented as a base-φ numeral using only the digits 0 and 1, and avoiding the digit sequence "11" ? this is called a standard form. A base-φ numeral that includes the digit sequence "11" can always be rewritten in standard form, using the algebraic properties of the base φ ― most notably that φ + 1 = φ ² . For instance, 11(φ) = 100(φ). Despite using an irrational number base, when using standard form, all on-negative integers have a unique representation as a terminating (finite) base-φ expansion. The set of numbers which possess a finite base-φ representation is the ring Z[1 + √5/2]; it plays the same role in this numeral systems as dyadic rationals play in binary numbers, providing a possibility to multiply.
Other numbers have standard representations in base-φ, with rational numbers having recurring representations. These representations are unique, except that numbers (mentioned above) with a terminating expansion also have a non-terminating expansion, as they do in base-10; for example, 1=0.99999….
Coach MMM, an Computer Science Professor who is also addicted to Mathematics, is extremely interested in GRB and now ask you for help to write a converter which, given an integer N in base-10, outputs its corresponding form in base-φ.

Input

There are multiple test cases. Each line of the input consists of one positive integer which is not larger than 10^9. The number of test cases is less than 10000. Input is terminated by end-of-file.

Output

For each test case, output the required answer in a single line. Note that trailing 0s after the decimal point should be wiped. Please see the samples for more details.

Sample Input

Sample Output

Hint

由于φ + 1 = φ ²，两边同乘φ ^k,得到φ ^k+1+φ ^k=φ ^k+2，说明只有有两位是1，就往前进一位。此外由φ + 1 = φ ²推到的2φ ²=φ ³+1，同理可知：φ ^k+3+φ ^k=2φ ^k+2，说明每一位的2都可以，由它前一位和它的后两位的1构成，这样就能将所有大于2的数降成1.再配合之前的，反复模拟便可得。由于当场没有估算这个数的长度，所以采用两个数组分别存了整数部分和小数部分。整体效率不是非常高，但是在短时间内做出来还是很高兴的。

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <queue>
#include <string>
#define inf 0x3fffffff
#define esp 1e-10
#define N 100

using namespace std;

int z[N], x[N], lenz, lenx;

bool judge ()
{
    if(z[0] && x[0])
        return 0;
    for (int i = 0; i < lenx; ++i)
        if (x[i] > 1 || (x[i] && x[i+1]))
            return 0;

    for (int i = 0; i < lenz; ++i)
        if (z[i] > 1 ||(z[i] ==1 && z[i+1] == 1))
            return 0;

    return 1;
}

void doz (int i)
{
    if (i == lenz-1)
        lenz++;
    int up = z[i] / 2;
    z[i] = z[i] & 1;
    z[i+1] += up;
    if (i >= 2)
        z[i-2] += up;
    else
    {
        if (lenx < 3 - i)
            lenx = 3 - i;
        x[1-i] += up;
    }
}

void dox (int i)
{
    if (i+3 > lenx)
        lenx = i + 3;
    int up = x[i] / 2;
    x[i] = x[i] & 1;
    x[i+2] += up;
    if (i == 0)
        z[0] += up;
    else
        x[i-1] += up;
}

void qt (int n)
{
    memset (z, 0, sizeof(z));
    memset (x, 0, sizeof(x));
    lenz = 1;
    lenx = 0;
    z[0] = n;
    while (!judge ())
    {
        for (int i = lenx-1; i >= 0; --i)
        {

            if (i == 0 && x[i] > 0 && x[i+1] > 0)
            {
                int up = min (x[i], x[i+1]);
                z[0] += up;
                x[0] -= up;
                x[1] -= up;
                continue;
            }
            else if (x[i] > 0 && x[i+1] > 0)
            {
                int up = min (x[i], x[i+1]);
                x[i-1] += up;
                x[i+1] -= up;
                x[i] -= up;
                continue;
            }
            if (x[i] > 1)
            {
                dox (i);
                continue;
            }


        }
        while(x[lenx-1] == 0)
            lenx--;
        for (int i = 0; i < lenz; ++i)
        {

            if (i == 0 && z[i] > 0 && x[0] > 0)
            {
                if (i == lenz-1)
                    lenz++;
                int up = min (z[i], x[0]);
                z[1] += up;
                z[0] -= up;
                x[0] -= up;
                continue;
            }
            else if (z[i] > 0 && z[i+1] > 0)
            {
                if (i+3 > lenz)
                    lenz = i + 3;
                int up = min (z[i], z[i+1]);
                z[i+2] += up;
                z[i+1] -= up;
                z[i] -= up;
                continue;
            }
            if (z[i] > 1)
            {
                doz(i);
                continue;
            }
        }
    }
    while(x[lenx-1] == 0)
        lenx--;
}

int main()
{
    //freopen ("test.txt", "r", stdin);
    int n;
    while (scanf ("%d", &n) != EOF)
    {
        qt (n);
        for (int i = lenz - 1; i >= 0; --i)
            printf ("%d", z[i]);
        if (lenx > 0)
            printf (".");
        for (int i = 0; i < lenx; ++i)
            printf ("%d", x[i]);
        printf ("\n");
    }
    return 0;
}

ACM学习历程——HDU4814 Golden Radio Base（数学递推）（12年成都区域赛）

标签：des style blog http io os ar for sp

原文地址：http://www.cnblogs.com/andyqsmart/p/4046563.html

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周排行

ACM学习历程——HDU4814 Golden Radio Base（数学递推） （12年成都区域赛）

ACM学习历程——HDU4814 Golden Radio Base（数学递推）（12年成都区域赛）