Fermat vs. Pythagoras
| Time Limit: 2000MS | Memory Limit: 10000K | |
| Total Submissions: 1493 | Accepted: 865 |
Description
Computer generated and assisted proofs and verification occupy a small niche in the realm of Computer Science. The first proof of the four-color problem was completed with the assistance of a computer program and current efforts in verification have succeeded in verifying the translation of high-level code down to the chip level.
This problem deals with computing quantities relating to part of
Fermat‘s Last Theorem: that there are no integer solutions of a^n + b^n =
c^n for n > 2.
Given a positive integer N, you are to write a program that computes
two quantities regarding the solution of x^2 + y^2 = z^2, where x, y,
and z are constrained(驱使)
to be positive integers less than or equal to N. You are to compute the
number of triples (x,y,z) such that x < y < z, and they are
relatively prime, i.e., have no common divisor(除数)
larger than 1. You are also to compute the number of values 0 < p
<= N such that p is not part of any triple (not just relatively prime
triples).
Input
The
input consists of a sequence of positive integers, one per line. Each
integer in the input file will be less than or equal to 1,000,000. Input
is terminated by end-of-file
Output
For
each integer N in the input file print two integers separated by a
space. The first integer is the number of relatively prime triples (such
that each component of the triple is <=N). The second number is the
number of positive integers <=N that are not part of any triple whose
components are all <=N. There should be one output line for each
input line.
Sample Input
10 25 100
Sample Output
1 4 4 9 16 27 题意：给定一个n,输出三元组(a,b,c)其中a，b，c两两互质，且a²+b²=c²,以及1~n中没有在任何一个三元组中出现过的数的个数。 这道题需要知道勾股数的性质。 首先，对于一组勾股数，①a与b的奇偶性不同，②c一定为奇数。 证明①：若a与b同为偶数，则c也为偶数，与a，b，c两两互质矛盾；若a与b同为奇数，c一定为偶数，设a=2*i+1，b=2*j+1，c=2*k -> a²+b²=c²->2*i²+2*i+2*j²+2*j+1=2*k²,这个式子是矛盾的。 证明②：a，b一奇一偶，显然。 然后将 a²+b²=c² 变形，a²=(c+b)*(c-b),容易发现c-b与c+b互质，所以c-b与c+b都是平方数，设x²=c-b，y²=c+b，得a=x*y，b=(y²-x²)/2,c=(y²+x²)/2. 易得x，y都为奇数,直接枚举x，y就好了。
1 #include <iostream>
2 #include <cstring>
3 #include <cstdio>
4 #include <cmath>
5 using namespace std;
6 const int maxn=1000010;
7 bool vis[maxn];
8 long long Gcd(long long a,long long b){
9 return b?Gcd(b,a%b):a;
10 }
11 int main(){
12 int n,m,ans,tot;
13 while(~scanf("%d",&n)){
14 m=(int)sqrt(n+0.5);ans=tot=0;
15 memset(vis,0,sizeof(vis));
16 for(int t=1;t<=m;t+=2)
17 for(int s=t+2;(s*s+t*t)/2<=n;s+=2)
18 if(Gcd(s,t)==1){
19 int a=s*t;
20 int b=(s*s-t*t)/2;
21 int c=(s*s+t*t)/2
22 ans++;
23 for(int k=1;k*c<=n;k++){
24 vis[k*a]=true;
25 vis[k*b]=true;
26 vis[k*c]=true;
27 }
28 }
29 for(int i=1;i<=n;i++)
30 if(!vis[i])
31 tot++;
32 printf("%d %d\n",ans,tot);
33 }
34 }