HDU 6441 Find Integer

不念不忘少年蓝@ 2022-05-15 07:04 134阅读 0赞

## 题目：[点击打开链接][Link 1] ##

## 题意：已知a^n+b^n=c^n，给出n和a，求b，c，如果无解输出−1 −1。  ##

## 分析：必须scanf才能过，时间卡的比较紧。 ##

## **费马大定理**  ##

## 1. a^n+b^n=c^n，n>2时无解。    2. 当a为奇数时，    a=2⋅k+1 ##

## c=k^2+(k+1)^2 ##

## b=c−1 ##

## 当 a 为偶数 ##

## a=2∗k+2 ##

## c=1+(k+1)^2 ##

## b=c−2 ##

## 手动列个方程就解出来了。 ##

## 代码： ##

#pragma GCC optimize(2)
    #pragma GCC optimize(3)
    #pragma GCC optimize(4)
    #pragma comment(linker, "/STACK:102400000,102400000")
    #include<unordered_map>
    #include<unordered_set>
    #include<algorithm>
    #include<iostream>
    #include<fstream>
    #include<complex>
    #include<cstdlib>
    #include<cstring>
    #include<cassert>
    #include<iomanip>
    #include<string>
    #include<cstdio>
    #include<bitset>
    #include<vector>
    #include<cctype>
    #include<cmath>
    #include<ctime>
    #include<stack>
    #include<queue>
    #include<deque>
    #include<list>
    #include<set>
    #include<map>
    using namespace std;
    #define pt(a) cout<<a<<endl
    #define debug test
    #define mst(ss,b) memset((ss),(b),sizeof(ss))
    #define rep(i,a,n) for (int i=a;i<=n;i++)
    #define per(i,a,n) for (int i=n-1;i>=a;i--)
    #define all(x) (x).begin(),(x).end()
    #define fi first
    #define se second
    #define SZ(x) ((int)(x).size())
    #define ll long long
    #define ull unsigned long long
    #define pb push_back
    #define mp make_pair
    #define inf 0x3f3f3f3f
    #define eps 1e-10
    #define PI acos(-1.0)
    typedef pair<int,int> PII;
    const ll mod = 1e9+7;
    const int N = 1e6+10;
    
    ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
    ll qp(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
    int to[4][2]={
       {-1,0},{1,0},{0,-1},{0,1}};
    
    int t,n,a,b,c;
    
    int main() {
        ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
        ///cin>>t;
        scanf("%d",&t);
        while(t--) {
            ///cin>>n>>a;
            scanf("%d%d",&n,&a);
            if(n>2 || n==0 ) printf("-1 -1\n");///cout<<-1<<" "<<-1<<endl;
            else {
                if(n==1) printf("%d %d\n",a,2*a);///cout<<a<<" "<<2*a<<endl;
                else if(a&1) b=(1+a*a)/2-1,c=b+1,printf("%d %d\n",b,c);///cout<<b<<" "<<c<<endl;
                else b=(1+a*a/4)-2,c=b+2,printf("%d %d\n",b,c);///cout<<b<<" "<<c<<endl;
            }
        }
        return 0;
    }

[Link 1]: http://acm.hdu.edu.cn/showproblem.php?pid=6441