Description：
people in USSS love math very much, and there is a famous math problem
give you two integers n,a,you are required to find
2
2
2 integers
b
,
c
b,c
b,c such that
a
n
+
b
n
=
c
n
a_{n}+b_{n}=c_{n}
an+bn=cn.
Input
one line contains one integer T ; ( 1 ≤ T ≤ 1000000 ) T;(1≤T≤1000000) T;(1≤T≤1000000)
next T lines contains two integers n , a ; ( 0 ≤ n ≤ 1000 , 000 , 000 , 3 ≤ a ≤ 40000 ) n,a;(0≤n≤1000,000,000,3≤a≤40000) n,a;(0≤n≤1000,000,000,3≤a≤40000)
Output
print two integers b , c b,c b,c if b , c b,c b,c exits;(￥1≤b,c≤1000,000,000￥);
else print two integers − 1 − 1 -1 -1 −1−1 instead.
Sample Input
1
2 3
Sample Output
4 5
题意：
每次两个整数
N
、
a
N、a
N、a，输出符合公式
a
n
+
b
n
=
c
n
a^n+b^n=c^n
an+bn=cn 的一对
b
b
b 和
c
c
c
这道题是我们暑假为了打区域赛网络赛训练的第一套题上面的，这个应该是签到题，当时还是yzq告诉我当
n
>
2
n>2
n>2 的时候无解，直接费马大定理。时间真快都一年。
- 当 n = 1 n = 1 n=1，即 a + b = c a + b = c a+b=c，已知 a a a ，令 b = 1 b = 1 b=1， c = a + 1 c = a + 1 c=a+1 即可
- 当 n = 2 n = 2 n=2，对于 x 2 + y 2 = z 2 x^2+y^2=z^2 x2+y2=z2 ,如果已知 x x x 那么就有 z 2 − y 2 = x 2 z^2−y^2=x^2 z2−y2=x2
- 因为 z > y z > y z>y，所以设 z = y + i z = y + i z=y+i,那么就有： x 2 − i 2 = 2 ∗ i ∗ y x^2−i^2=2∗i∗y x2−i2=2∗i∗y
- 只要枚举 i i i ，判断（ x 2 − i 2 x^2 - i^2 x2−i2） % 2 ∗ i \% 2*i %2∗i是否可以整除即可。
AC代码：
#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <stack>
#include <queue>
using namespace std;
#define sd(n) scanf("%d", &n)
#define sdd(n, m) scanf("%d%d", &n, &m)
#define sddd(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define pd(n) printf("%d\n", n)
#define pc(n) printf("%c", n)
#define pdd(n, m) printf("%d %d", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n, m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld", &n)
#define sldd(n, m) scanf("%lld%lld", &n, &m)
#define slddd(n, m, k) scanf("%lld%lld%lld", &n, &m, &k)
#define sf(n) scanf("%lf", &n)
#define sc(n) scanf("%c", &n)
#define sff(n, m) scanf("%lf%lf", &n, &m)
#define sfff(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define ss(str) scanf("%s", str)
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
#define mem(a, n) memset(a, n, sizeof(a))
#define debug(x) cout << #x << ": " << x << endl
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define mod(x) ((x) % MOD)
#define gcd(a, b) __gcd(a, b)
#define lowbit(x) (x & -x)
typedef pair<int, int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int MOD = 1e9 + 7;
const double eps = 1e-9;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
inline int read()
{
int ret = 0, sgn = 1;
char ch = getchar();
while (ch < '0' || ch > '9')
{
if (ch == '-')
sgn = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
{
ret = ret * 10 + ch - '0';
ch = getchar();
}
return ret * sgn;
}
inline void Out(int a) //Êä³öÍâ¹Ò
{
if (a > 9)
Out(a / 10);
putchar(a % 10 + '0');
}
ll gcd(ll a, ll b)
{
return b == 0 ? a : gcd(b, a % b);
}
ll lcm(ll a, ll b)
{
return a * b / gcd(a, b);
}
///快速幂m^k%mod
ll qpow(ll a, ll b, ll mod)
{
if (a >= mod)
a = a % mod + mod;
ll ans = 1;
while (b)
{
if (b & 1)
{
ans = ans * a;
if (ans >= mod)
ans = ans % mod + mod;
}
a *= a;
if (a >= mod)
a = a % mod + mod;
b >>= 1;
}
return ans;
}
// 快速幂求逆元
int Fermat(int a, int p) //费马求a关于b的逆元
{
return qpow(a, p - 2, p);
}
///扩展欧几里得
int exgcd(int a, int b, int &x, int &y)
{
if (b == 0)
{
x = 1;
y = 0;
return a;
}
int g = exgcd(b, a % b, x, y);
int t = x;
x = y;
y = t - a / b * y;
return g;
}
///使用ecgcd求a的逆元x
int mod_reverse(int a, int p)
{
int d, x, y;
d = exgcd(a, p, x, y);
if (d == 1)
return (x % p + p) % p;
else
return -1;
}
///中国剩余定理模板0
ll china(int a[], int b[], int n) //a[]为除数，b[]为余数
{
int M = 1, y, x = 0;
for (int i = 0; i < n; ++i) //算出它们累乘的结果
M *= a[i];
for (int i = 0; i < n; ++i)
{
int w = M / a[i];
int tx = 0;
int t = exgcd(w, a[i], tx, y); //计算逆元
x = (x + w * (b[i] / t) * x) % M;
}
return (x + M) % M;
}
#include <cstdio>
typedef long long ll;
int main()
{
int t;
sd(t);
while (t--)
{
ll n, a, b, c;
sldd(n, a);
if (n == 1)
{
pldd(1, 1 + a);
}
else if (n == 2)
{
if (a % 2 == 1)
{
ll res = (a - 1) / 2;
b = 2 * res * res + 2 * res;
c = b + 1;
pldd(b, c);
}
else
{
ll res = a / 2 - 1;
b = res * res + 2 * res;
c = b + 2;
pldd(b, c);
}
}
else
printf("-1 -1 \n");
}
return 0;
}