链接：https://ac.nowcoder.com/acm/contest/1083/C来源：牛客网

题目描述

给出直角三角形其中一条边的长度n，你的任务是构造剩下的两条边，使这三条边能构成一个直角三角形。

输入描述:

一个整数n。

输出描述:

另外两条边b,c。答案不唯一，只要输出任意一组即为合理，如果无法构造请输出-1。

示例1

输入

复制

3

输出

复制

4 5

示例2

输入

复制

4

输出

复制

3 5

备注:

0<=n<=1e91<=b,c<=1e18n,b,c均为整数

思路：

通过一下程序打表，可以发现规律：

当\(n>=3\) 时，一定有答案。

当n为奇数时，一定有a,b 满足 a=b-1 ，\(a*a+n*n=b*b\)

当n为偶数时，一定有a,b 满足 a=b-2 ，\(a*a+n*n=b*b\)

把b用a表示就是一个一元一次方程，求解即可。

打表程序：

#include <bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define du2(a,b) scanf("%d %d",&(a),&(b))
#define du1(a) scanf("%d",&(a));
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {a %= MOD; if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}

inline void getInt(int* p);
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
string S(ll n) {stringstream ss; string s; ss << n; ss >> s; return s;}
ll N(string s) {stringstream ss; ll n; ss << s; ss >> n; return n;}
string rm0(string s) { //  去除前导0函数
    int i;
    for (i = 0; i < s.size() - 1; i++)
        if (s[i] != '0')
            break;
    return s.substr(i);
}
string ADD(string s, string t) { // 字符串整数相加，返回一个字符串
    if (s.size() < t.size())swap(s, t); s = '0' + s;
    reverse(t.begin(), t.end()); while (s.size() > t.size())t += '0'; reverse(t.begin(), t.end()); int c = 0;
    for (int i = s.size(); i >= 0; i--)
    {
        if (c) {
            if (s[i] == '9') {
                s[i] = '0'; c = 1;
            } else {
                s[i] = (char)(s[i] + 1); c = 0;
            }
        }
        int sum = (int)s[i] + (int)t[i] - '0' * 2;
        if (sum >= 10) {
            s[i] = (char)(sum - 10 + '0'); c = 1;
        }
        else
            s[i] = (char)(sum + '0');
    }
    return rm0(s);
}
bool cmp(string s, string t) {
    // s>=t 返回1
    // s<t 返回0
    if (s.size() != t.size())
        return s.size() > t.size();
    for (int i = 0; i < s.size(); i++)
        if (s[i] != t[i])
            return s[i] > t[i];
    return 1;
}

// 字符串减法，不带负号
// int x[MAXN], y[MAXN];
// string minuss(string a, string b)
// {
//     int la = a.length(), lb = b.length(), lc, i;
//     memset(y, 0, sizeof(y)); lc = max(la, lb); string ans;
//     for (i = 0; i < la; i++)x[la - i - 1] = a[i] - 48;
//     for (i = 0; i < lb; i++)y[lb - i - 1] = b[i] - 48;
//     for (i = 0; i < lc; i++)
//     {
//         x[i] -= y[i];
//         if (x[i] < 0)x[i + 1]--, x[i] += 10;
//     }
//     while (!x[lc - 1] && lc > 1)lc--;
//     for (i = lc - 1; i + 1; i--)ans.push_back(x[i] + 48);
//     return ans;
// }


string add(string a, string b)
{
    // 二进制带补位加法
    int len1 = a.size(), len2 = b.size();
    if (len1 > len2)swap(a, b), swap(len1, len2);
    reverse(a.begin(), a.end()); reverse(b.begin(), b.end());
    for (int i = 0; i < len2 - len1; i++)a += "0";
    int f = 0;
    string ans = "";
    for (int i = 0; i < len2; i++) {
        int k = a[i] - '0' + b[i] - '0' + f; f = 0;
        if (k >= 2)k %= 2, f = 1;
        if (k == 0)ans += "0";
        else ans += "1";
    }
    if (f)ans += "1";
    reverse(ans.begin(), ans.end());
    return ans;
}
int a[maxn];
set<string> st;
int main()
{
    //freopen("D:\\code\\text\\input.txt","r",stdin);
    //freopen("D:\\code\\text\\output.txt","w",stdout);
    int n;
    cin >> n;

    repd(i, 1, n)
    {
        repd(j, 1, n)
        {
            repd(k, 1, n)
            {
                if (i * i + j * j == k * k || (j * j + k * k) == i * i)
                {
                    int g = 1;
                    a[0] = i / g;
                    a[1] = j / g;
                    a[2] = k / g;
                    sort(a, a + 3);
                    string s = S(a[0]) + S(a[1]) + S(a[2]);
                    if (st.count(s) == 0)
                    {
                        st.insert(s);
                        cout << a[0] << " " << a[1] << " " << a[2] << endl;
                    }
                }
            }
        }
    }
    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}

　ＡＣ代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define du2(a,b) scanf("%d %d",&(a),&(b))
#define du1(a) scanf("%d",&(a));
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {a %= MOD; if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}

inline void getInt(int* p);
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/

int main()
{
    //freopen("D:\\code\\text\\input.txt","r",stdin);
    //freopen("D:\\code\\text\\output.txt","w",stdout);
    ll n;
    cin >> n;
    if (n <= 2)
    {
        cout << -1 << endl;
    } else
    {
        if (n & 1)
        {
            cout << ((n * n - 1) >> 1) << " " << (n * n - 1ll >> 1) + 1 << endl;
        } else
        {
            cout << (n * n - 4ll) / 4ll << " " << (n * n - 4ll) / 4ll + 2 << endl;
        }
    }
    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}