【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 secondtypedef 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;}