poj 1305
2015-11-04 22:42
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Fermat vs. Pythagoras
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
Sample Output
Source
Duke Internet Programming Contest 1991,UVA 106
暴力枚举就好
View Code
Time Limit: 2000MS | Memory Limit: 10000K | |
Total Submissions: 1450 | Accepted: 846 |
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
#include <cstdio> #include <cstring> #include <iostream> #include <stack> #include <queue> #include <map> #include <algorithm> #include <vector> #include <cmath> using namespace std; const int maxn = 1000001; typedef long long LL; bool flag[maxn]; int gcd(int a,int b) { if(b == 0) return a; else return gcd(b,a%b); } void solve(int t) { int temp,m,i,j,k,n,ans1,ans2,x,y,z; memset(flag,0,sizeof(flag)); temp = sqrt(t+0.0); ans1 = ans2 = 0; for( i=1;i<=temp;i++){ for(j = i+1;j<=temp;j++){ if(i*i + j*j > t) break; if((i%2) != (j%2)){ if(gcd(i,j) == 1){ x = j*j - i*i; y = 2*i*j; z = j*j + i*i; ans1++; for( k=1;;k++){ if( k*z > t) break; flag[k*x] = 1; flag[k*y] = 1; flag[k*z] = 1; } } } } } for(int i=1;i<=t;i++){ if(!flag[i]) ans2++; } printf("%d %d\n",ans1,ans2); } int main() { int n; while(scanf("%d",&n)!=EOF){ solve(n); } return 0; }
View Code
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