【9308】极值问题

Time Limit: 10 second
Memory Limit: 2 MB

问题描述
已知m，n为整数，且满足下列两个条件：
1、m、n∈{1，2…,k},即1≤m，n≤k；
2、(n^2-m*n-m^2)^2=1。
编程输入正整数k(1≤k≤10^9),求一组满足上述两个条件的m，n，并且使得n^2+m^2的值最大。
Input

输入正整数k

Output

输出满足最大值的m和n

Sample Input

100

Sample Output

m=55
n=89

【题目链接】:http://noi.qz5z.com/viewtask.asp?id=9308

【题解】

(n^2-m*n-m^2)^2=1
(m^2+m*n-n^2)^2=1
(m^2+2*m*n+n^2-m*n-2*n^2)^2=1
(m^2+2*m*n+n^2-m*n-n^2-n^2)^2=1
(m^2+2*m*n+n^2-n*(m+n)-n^2)^2=1
((m+n)^2-n*(m+n)-n^2)^2=1
则令n’=(m+n),m’=n;
(n’^2-m’*n’-m’^2)^2=1
所以如果n和m是符合要求的解；
则n=(m+n)和m=n也是方程的解；
而由观察可知n=1,m=1是方程的解
所以n=2,m=1也是方程的解；
则n=3;m=2也是方程的解。。
就是斐波那契数列了；
递推下就好;

【完整代码】

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <set>
#include <map>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
#include <stack>
#include <string>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

void rel(LL &r)
{
    r = 0;
    char t = getchar();
    while (!isdigit(t) && t!='-') t = getchar();
    LL sign = 1;
    if (t == '-')sign = -1;
    while (!isdigit(t)) t = getchar();
    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
    r = r*sign;
}

void rei(int &r)
{
    r = 0;
    char t = getchar();
    while (!isdigit(t)&&t!='-') t = getchar();
    int sign = 1;
    if (t == '-')sign = -1;
    while (!isdigit(t)) t = getchar();
    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
    r = r*sign;
}

//const int MAXN = x;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);

LL sqr(LL x)
{
    return x*x;
}

LL k;

int main()
{
   // freopen("F:\rush.txt","r",stdin);
    cin >> k;
    LL a = 1,b = 1,c;
    while ((a+b)<=k)
    {
        c = a+b;
        a = b;
        b = c;
    }
    cout << "m="<<a<<endl;
    cout << "n="<<b<<endl;
    return 0;
}