NBUT 1224 - Happiness Hotel【佩爾方程】

Happiness Hotel

Time Limit:1000MS     Memory Limit:131072KB

Description：

The life of Little A is good, and, he managed to get enough money to run a hotel. The best for him is that he need not go to work outside, just wait for the money to go into his pocket. Little A wants everything to be perfect, he has a wonderful plan that he will keep one most beautiful reception whose size is 1()(which means the reception is 1 square meter). There are other k rooms that have the same area, and the area is x^2(), x is an integer; Little A wants his hotel to be a square. Little A is a good thinker, but not a good maker. As his poor performance on math, he cannot calculate the least area needed to build such a hotel of his will. Now, this task belongs to you, solve this problem to make Little A’s dream of Happy Hotel come true. Please be careful, the whole area should only contain k rooms, and the reception, there should not be any vacant place.

Input：

There are several test cases. 

Each case contains only one integer k(1<=k<=1000) ,the number of rooms the hotel should have in one line. 

Proceed to the end of file.

Output：

Output one integer d, means the hotel’s area is d^2(If there is no answer, output “no solution”) .The output of one test case occupied exactly one line.

Sample Input：

1

2

3

Sample Output：

no solution

3

2

題意：

輸入一個整數k，求解使得方程 ans ^ 2 = k * n ^ 2 + 1 成立的 ans 最小值(n是大於1的整數)

算法思路：

從題意可以看出是佩爾方程,直接調用佩爾方程模板解決

代碼實現：

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;

int can[1005] = {0};
int a[10005][605]= {0};
int x[6005], y[6005], t[6005];
int h1,h2;
int bb, ee, xx, yy, c, n;

void Pell(int ji,int many,int ma,int kk)
{
    if (ji < kk)
        Pell(ji + 1, a[ma][(ji-1)%a[ma][600]+1], ma, kk);
    else
    {
        h1 = 1;
        h2 = 1;
        x[1] = many;
        y[1] = 1;
        return;
    }
    for (int i = 1; i <= h1; i++)
        t[i] = x[i];
    for (int i = 1; i <= h2; i++)
        x[i] = y[i];
    for (int i = 1; i <= h1; i++)
        y[i] = t[i];
    c = h1;
    h1 = h2;
    h2 = c;
    for (int i = 1; i <= h2; i++)
        if (i <= h1)
            x[i] += many * y[i];
        else
            x[i] = many * y[i];
    if (h2 > h1)
        h1 = h2;
    for (int i = 1; i < h1; i++)
        if (x[i] >= 10)
        {
            x[i+1] += x[i] / 10;
            x[i] %= 10;
        }
    while(x[h1] >= 10)
    {
        x[h1+1] = x[h1] / 10;
        x[h1] %= 10;
        h1++;
    }
    x[0] = h1;
}

void solve()
{
    int i, j;
    for (j = 1; j <= 31; j++)
        can[j*j] = true;
    for(i = 1; i <= 1000; i++)
    {
        if(!can[i])
        {
            a[i][600]=1;
            bb = 1;
            ee = (int)sqrt((double)i);
            a[i][0] = ee;
            ee =- ee;
            xx = bb;
            yy = ee;
            xx =- yy;
            yy = i - yy * yy;
            n=0;
            while((xx - yy) * (xx - yy) < i || xx >= 0)
            {
                xx -= yy;
                n++;
            }
            a[i][1] = n;
            c = xx, xx = yy, yy = c;
            while(xx != bb || yy != ee)
            {
                a[i][600]++;
                c = xx;
                xx =- yy;
                yy = i - yy * yy;
                yy = yy / c;
                n = 0;
                while ((xx - yy) * (xx - yy) < i || xx >= 0)
                {
                    xx -= yy;
                    n++;
                }
                a[i][a[i][600]] = n;
                c = xx, xx = yy, yy = c;
            }
        }
    }
}

int main()
{
    int k;
    solve();
    while(scanf("%d",&k)!=EOF)///ans^2 = k * n ^ 2 + 1,求ans的最小值(k <= 1000)
    {
        if(!can[k])
        {
            if(a[k][600] % 2)
                Pell(1, a[k][0], k, a[k][600]*2);
            else
                Pell(1, a[k][0], k, a[k][600]);
            for (int j = x[0]; j >= 1; j--)
                printf("%d",x[j]);
            printf("\n");
        }
        else
            printf("no solution\n");
    }
    return 0;
}