ZOJ Problem Set - 3673 1729

1729 Time Limit: 3 Seconds       Memory Limit: 65536 KB

1729 is the natural number following 1728 and preceding 1730. It is also known as the Hardy-Ramanujan number after a famous anecdote of the British mathematician G. H. Hardyregarding a hospital visit to the Indian mathematician Srinivasa Ramanujan. In Hardy's words:

I remember once going to see him when he was ill at Putney. I had ridden in taxi cab number 1729 and remarked that the number seemed to me rather a dull one, and that I hoped it was not an unfavorable omen. "No," he replied, "it is a very interesting number; it is the smallest number expressible as the sum of two (positive) cubes in two different ways."

The two different ways are these: 1729 = 1³ + 12³ = 9³ + 10³

Now your task is to count how many ways a positive number can be expressible as the sum of two positive cubes in. All the numbers in this task can be expressible as the sum of two positive cubes in at least one way.

Input

There're nearly 20,000 cases. Each case is a positive integer in a single line. And all these numbers are greater than 1 and less than 2⁶⁴.

Output

Please refer to the sample output. For each case, you should output a line. First the number of ways n. Then followed by n pairs of integer, (a_i,b_i), indicating a way the given number can be expressible as the sum of a_i's cube and b_i's. (a_i≤ b_i, and a₁< a₂< ...< a_n)

Sample Input

9
4104
2622104000
21131226514944
48988659276962496

Sample Output

1 (1,2)
2 (2,16) (9,15)
3 (600,1340) (678,1322) (1020,1160)
4 (1539,27645) (8664,27360) (11772,26916) (17176,25232)
5 (38787,365757) (107839,362753) (205292,342952) (221424,336588) (231518,331954)

//zoj 1729

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>


using namespace std;
#define Z 6000000
#define maxn 2642246
typedef unsigned long long ull;




ull prime[Z], np;
bool vis[Z];
ull zys[1010],dycishu[1010];
ull n;
int top;
int len;
struct cc
{
    ull a;
    ull b;
} ans[10000];




int cmp(cc x,cc y)
{
return x.a<y.a;
}








void primebiao()
{
    int i,j;
    np = 0;
    for(i = 2; i<Z; i++)
    {
        if(!vis[i]) prime[np++] = i;
        for(int j = 0; j<np &&  prime[j]*i<Z; j++)
        {
            vis[prime[j]*i] = 1;
            if(i%prime[j]==0) break;
        }
    }
}


ull div(ull m)   //找出n的所有的因子
{
int i;
    ull k;
    len = 0;
    for(i = 0; i<np && prime[i]*prime[i]<=m; i++)
    {
        if(m%prime[i] == 0)
        {
            zys[len] = prime[i];  //保存因子
            k = 1;
            m/=prime[i];
            while(m%prime[i] == 0)
            {
                k++;
                m/=prime[i];
            }
            dycishu[len++] = k;   //每个因子的次数
        }
    }
    if(m>1)    //本身是质数的时候
    {
        zys[len] = m;
        dycishu[len++] = 1;
    }


}




ull quickpow(ull m,ull n)
{
    ull b = 1;
    while (n > 0)
    {
          if (n & 1)
             b = b*m;
          n = n >> 1 ;
          m = m*m;
    }
    return b;
}


ull judge(ull derta)  //要求整数解，b^2-4*a*c必须能开出整数
{
    ull x = sqrt(1.0*derta);
    if(x*x ==derta) return x;
    return Z;
}


int panduanchongfu(ull x,ull y)  //看这组解是否已经存在
{
    for(int i=0; i<top; i++)
        if(ans[i].a==x&&ans[i].b==y)
            return 1;
    return 0;
}






void work(ull t)   //找出解..
{
ull x1,x2;
    ull k = n/t;
    ull r = t*t-k; // a+b=t,(t*t-n/t)/3=p,  a^2-t*a+p=0;
    if(r>0 && r%3!=0) return;
    r/=3;
    ull dis = t*t-4*r;
    if(dis<0) return ;
    ull c = judge(dis);
    if(c == Z) return;


    if((t+c)%2 == 0)
    {
        x1 = (t+c)/2;
        x2 = t-x1;
        if(x1>x2) swap(x1,x2);
        if(x1>0&&x1<t&&x1<maxn&&x2<maxn&&x2>0&&x2<t&&!panduanchongfu(x1,x2))
        {
            ans[top].a=x1;
            ans[top++].b=x2;
        }
    }
    if((t-c)>0 && (t-c)%2 == 0)
    {
        x1 = (t-c)/2;
        x2 = t-x1;
        if(x1>x2) swap(x1,x2);
        if(x1>0&&x1<t&&x1<maxn&&x2<maxn&&x2>0&&x2<t&&!panduanchongfu(x1,x2))
        {
            ans[top].a=x1;
            ans[top++].b=x2;
        }
    }
}








void dfs(ull m,ull sum)
{
    work(sum);
    if(m>=len)  return ;   //len是n的因子的个数，m是代表第m个因子
    for(int i = 0; i<=dycishu[m]; i++)
        dfs(m+1,sum*quickpow(zys[m],i));
}








int main()
{
    //freopen("1.txt","r",stdin);
primebiao();   //打素数表
while(scanf("%llu",&n)!=EOF)
{
div(n);
//for(int i=0;i<len;i++)
// cout<<zys[i]<<' '<<dycishu[i]<<endl;
top=0;
dfs(0,1);
sort(ans,ans+top,cmp);
printf("%d",top);


         for(int i = 0; i<top; i++)
            printf(" (%llu,%llu)",ans[i].a,ans[i].b);
        printf("\n");




}
return 0;
}