POJ 1305 Fermat vs. Pythagoras (本原勾股数组)

Fermat vs. Pythagoras

Time Limit: 2000MS Memory Limit: 10000KTotal Submissions: 1467 Accepted: 850

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 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 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

1025100

Sample Output

1 4

4 9

16 27

勾股数组就是能形成a^2+b^2=c^2的一组数（a,b,c），根据规律可发现，a,b,c三个数是互素的，对于任意勾股数组

a=s*t;

b=(s*s-t*t)/2;

c=(s*s+t*t)/2

其中s>t且s和t互素

题意：寻找n以内（包含n）的勾股数组的数量还有不是勾股数组的数的数量

思路：根据勾股数组的定义来枚举，为了省时，进行筛选，中间需要判断c就行了，根据定义可知，c是最大的

再者因为枚举过程中s*s+t*t>=2*s*t，那么c的值就是最大的

总结：刚开始想着枚举a,b,c，但是看了100w的数据量，还是算了，想到定义中可以根据s和t来求得a,b,c，那么枚举s和t就行了

ac代码：

#include<stdio.h>#include<math.h>#include<string.h>#include<stack>#include<queue>#include<vector>#include<iostream>#include<algorithm>#define MAXN 1010000#define LL long long#define ll __int64#define INF 0xfffffff#define mem(x) memset(x,0,sizeof(x))#define PI acos(-1)#define mod 1000000007using namespace std;int v[MAXN];ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}int main(){ll t,n,m,i,j,num;int cas=0;while(scanf("%I64d",&n)!=EOF){ll cnt=0;mem(v);for(i=1;i<=n;i++){for(j=i+1;j<=n;j++){if(gcd(i,j)!=1)continue;ll a=i*j;ll b=(j*j-i*i)/2;ll c=(i*i+j*j)/2;//printf("a=%I64d b=%I64d c=%I64d\n",a,b,c);if(a==c||b==c||a==b)continue;if(c>n)break;if(a*a+b*b==c*c)cnt++;elsecontinue;for(ll k=1;k*c<=n;k++)//对于勾股数组的倍数进行筛选{v[a*k]=1;v[b*k]=1;v[c*k]=1;}}}ll ans=0;for(i=1;i<=n;i++)if(!v[i])ans++;printf("%I64d %I64d\n",cnt,ans);}return 0;}