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
Duke Internet Programming Contest 1991,UVA 106
暴力枚举就好
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 }