【Codeforces 707C】Pythagorean Triples(找规律)
2016-08-21 18:25
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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; if(n>1&&n&1) cout<<n/2<<" "<<n/2+1; else if(n>4&&n%4==0) cout<<n/4-1<<" "<<n/4+1; else printf("-1"); }
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