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poj 1305

时间:2015-11-05 00:21:23      阅读:252      评论:0      收藏:0      [点我收藏+]

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Fermat vs. Pythagoras
Time Limit: 2000MS   Memory Limit: 10000K
Total Submissions: 1450   Accepted: 846

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

10
25
100

Sample Output

1 4
4 9
16 27

Source

暴力枚举就好
技术分享
 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 #include <stack>
 5 #include <queue>
 6 #include <map>
 7 #include <algorithm>
 8 #include <vector>
 9 #include <cmath>
10 
11 using namespace std;
12 
13 const int maxn = 1000001;
14 
15 typedef long long LL;
16 
17 bool flag[maxn];
18 
19 int gcd(int a,int b)
20 {
21     if(b == 0) return a;
22     else return gcd(b,a%b);
23 }
24 void solve(int t)
25 {
26     int temp,m,i,j,k,n,ans1,ans2,x,y,z;
27     memset(flag,0,sizeof(flag));
28     temp = sqrt(t+0.0);
29     ans1 = ans2 = 0;
30     for( i=1;i<=temp;i++){
31         for(j = i+1;j<=temp;j++){
32             if(i*i + j*j > t) break;
33             if((i%2) != (j%2)){
34                 if(gcd(i,j) == 1){
35                     x = j*j - i*i;
36                     y = 2*i*j;
37                     z = j*j + i*i;
38                     ans1++;
39                     for( k=1;;k++){
40                         if( k*z > t) break;
41                         flag[k*x] = 1;
42                         flag[k*y] = 1;
43                         flag[k*z] = 1;
44                     }
45                 }
46             }
47         }
48     }
49     for(int i=1;i<=t;i++){
50         if(!flag[i]) ans2++;
51     }
52     printf("%d %d\n",ans1,ans2);
53 }
54 int main()
55 {
56     int n;
57     while(scanf("%d",&n)!=EOF){
58         solve(n);
59     }
60     return 0;
61 }
View Code

 

poj 1305

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原文地址:http://www.cnblogs.com/lmlyzxiao/p/4937697.html

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