Codeforces Round #368 (Div. 2)（C. Pythagorean Triples 勾股数规律）

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题目链接

给出一个数，输出2个数，使得这三个数是勾股数

形如2n,n^2-1,n^2+1可以组合成勾股数，这是偶数的情况
奇数的时候看个例子
3, 4 , 5 | 4 = (1+3) * 1
5,12,13 | 12 = (1+5)*2
7,24,25 | 24 = (1+7)*3
9,40,41 | 40 = (1+9)*4
…
第二列数就是第一列数在以3为首项的等差数列中的位置*（该数+1）

#include<bits/stdc++.h>using namespace std;#define cl(a,b) memset(a,b,sizeof(a))#define fastIO ios::sync_with_stdio(false);cin.tie(0);#define LL long long#define pb push_back#define gcd __gcd#define For(i,j,k) for(int i=(j);i<k;i++)#define lowbit(i) (i&(-i))#define _(x) printf("%d\n",x)const int maxn = 3e6+10;const int inf  = 1 << 28;LL mx = 1e18;bool flag = false;void out(LL a,LL b) {    if(a>mx||b>mx||a<=0||b<=0) {        //printf("-1\n");        return;    } else {        printf("%lld %lld\n",a,b);        flag = true;    }}void work(LL n) {    flag = false;    if(n%3==0) {        out(n/3*4,n/3*5);    }    if(flag) return ;    if(n%4==0) {        out(n/4*3,n/4*5);    }    if(flag) return ;    if(n%5==0) {        out(n/5*3,n/5*4);    }    if(flag) return ;    if(n%2==1) {        LL t = (n-1)/2;        LL a = t*(n+1);        LL b = a+1;        out(a,b);    }    if(flag) return ;    if(n%2==0){        LL t = n/2;        LL a = t*t-1;        LL b = t*t+1;        out(a,b);    }    if(flag) return ;    puts("-1");}int main() {    LL n;    scanf("%lld",&n);    work(n);    return 0;}

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