zoj 3673 1729

標籤：style   blog   color   io   os   ar   for   sp   div   

1729Time 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. Hardy regarding 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, (ai,bi), indicating a way the given number can be expressible as the sum of ai‘s cube and bi‘s. (ai≤ bi, and a1< a2< ...< an)

Sample Input

9410426221040002113122651494448988659276962496

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)

Hint

Although most numbers cannot be expressible as the sum of two positive cubes, the vast majority of numbers in this task can be expressible as the sum of two positive cubes in two or more ways. 

 

  1 #include<iostream>  2 #include<stdio.h>  3 #include<cstring>  4 #include<math.h>  5 #include<cstdlib>  6 #include<algorithm>  7 using namespace std;  8 typedef unsigned long long LL;  9 const int maxn = 2642246+2; 10  11 struct node 12 { 13     LL x,y; 14 } tom[10002]; 15 int tlen; 16 LL prime[192728]; 17 int len; 18 bool s[maxn]; 19 void init() 20 { 21     memset(s,false,sizeof(s)); 22     len = 0; 23     for(int i=2; i<maxn; i++) 24         if(s[i]==false) 25         { 26             prime[++len]=i; 27             for(int j=i+i; j<maxn; j=j+i) 28                 s[j]=true; 29         } 30 } 31 LL fac[5001],num[5001]; 32 int flen; 33 void Euler(LL n) 34 { 35     int i,count; 36     flen = 0; 37     for(i=1; prime[i]*prime[i]<=n; i++) 38     { 39         if(n%prime[i]==0) 40         { 41             count = 0; 42             while(n%prime[i]==0) 43             { 44                 n=n/prime[i]; 45                 count++; 46             } 47             fac[++flen]=prime[i]; 48             num[flen]=count; 49         } 50     } 51     if(n!=1) 52     { 53         fac[++flen]=n; 54         num[flen]=1; 55     } 56 } 57  58 LL Q[50050]; 59 int qlen; 60 void solve() 61 { 62     Q[0]=1; 63     qlen = 0; 64     for(int i=1; i<=flen; i++) 65     { 66         int k; 67         int s=0; 68         for(int j=1; j<=num[i]; j++) 69         { 70             k = qlen; 71             for(; s<=k; s++) 72                 Q[++qlen]=Q[s]*fac[i]; 73         } 74     } 75 } 76 int main() 77 { 78     LL n; 79     int T=0; 80     init(); 81     while(scanf("%llu",&n)>0) 82     { 83         Euler(n); 84         solve(); 85         sort(Q+1,Q+1+qlen); 86         tlen=0; 87         int NUM=0; 88         for(int i=1; i<=qlen; i++) 89         { 90             LL y = (Q[i]*Q[i]-n/Q[i])/3; 91             LL x = Q[i]; 92             if(x*x>=4*y) 93             { 94                 LL ss = (LL)sqrt((x*x-4*y)*1.0); 95                 LL ans1 = (x-ss)/2; 96                 LL ans2 = (x+ss)/2; 97                 if(ans1==0||ans2==0)continue; 98                 if(ans1*ans1*ans1+ans2*ans2*ans2==n) 99                 {100                     tom[++tlen].x=ans1;101                     tom[tlen].y=ans2;102                     NUM++;103                 }104             }105         }106         printf("%d",NUM);107         for(int i=1; i<=tlen; i++)108             printf(" (%llu,%llu)",tom[i].x,tom[i].y);109         printf("\n");110     }111     return 0;112 }

 

zoj 3673 1729