Katya studies in a fifth grade. Recently her class studied right triangles and the Pythagorean theorem. It appeared, that there are triples of positive integers such that you can construct a right triangle with segments of lengths corresponding to tr…
题目链接: C. Pythagorean Triples time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Katya studies in a fifth grade. Recently her class studied right triangles and the Pythagorean theorem. It appea…
Pythagorean Triples 题目链接: http://codeforces.com/contest/707/problem/C Description Katya studies in a fifth grade. Recently her class studied right triangles and the Pythagorean theorem. It appeared, that there are triples of positive integers such th…
Pythagorean Triples time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Katya studies in a fifth grade. Recently her class studied right triangles and the Pythagorean theorem. It appeared, that…
C. Pythagorean Triples time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Katya studies in a fifth grade. Recently her class studied right triangles and the Pythagorean theorem. It appeared, t…
C. Pythagorean Triples 题目连接: http://www.codeforces.com/contest/707/problem/C Description Katya studies in a fifth grade. Recently her class studied right triangles and the Pythagorean theorem. It appeared, that there are triples of positive integers…
Description Katya studies in a fifth grade. Recently her class studied right triangles and the Pythagorean theorem. It appeared, that there are triples of positive integers such that you can construct a right triangle with segments of lengths corresp…
Description Katya studies in a fifth grade. Recently her class studied right triangles and the Pythagorean theorem. It appeared, that there are triples of positive integers such that you can construct a right triangle with segments of lengths corresp…
一边长为a的直角三角形,a^2=c^2-b^2.可以发现1.4.9.16.25依次差3.5.7.9...,所以任何一条长度为奇数的边a,a^2还是奇数,那么c=a^2/2,b=c+1.我们还可以发现,1.4.9.16.25.36各项差为8.12.16.20,偶数的平方是4的倍数,那么c=a^2/4-1,b=a^2/4+1. #include <iostream> using namespace std; int main() { long long n; cin>>n; n*=n;…
