Java实现蓝桥杯历届试题斐波那契

喜欢ヅ旅行 2023-05-28 11:26 13阅读 0赞

## 试题 历届试题 斐波那契 ##

资源限制  
时间限制：1.0s 内存限制：256.0MB  
问题描述  
　　斐波那契数列大家都非常熟悉。它的定义是：

f(x) = 1 … (x=1,2)  
　　f(x) = f(x-1) + f(x-2) … (x>2)

对于给定的整数 n 和 m，我们希望求出：  
　　f(1) + f(2) + … + f(n) 的值。但这个值可能非常大，所以我们把它对 f(m) 取模。  
　　公式如下  
![在这里插入图片描述][20200326095424227.png]

但这个数字依然很大，所以需要再对 p 求模。  
输入格式  
　　输入为一行用空格分开的整数 n m p (0 < n, m, p < 10^18)  
输出格式  
　　输出为1个整数，表示答案  
样例输入  
2 3 5  
样例输出  
0  
样例输入  
15 11 29  
样例输出  
25

import java.math.BigInteger;
    import java.util.Scanner;
    
    public class 斐波那契 { 
    	static BigInteger[][] cal_fm = {  {  new BigInteger("1"), new BigInteger("1") },
    			{  new BigInteger("1"), new BigInteger("0") } };
    	static BigInteger[][] cal_sum = {  {  new BigInteger("2"), new BigInteger("0"), new BigInteger("-1") },
    			{  new BigInteger("1"), new BigInteger("0"), new BigInteger("0") },
    			{  new BigInteger("0"), new BigInteger("1"), new BigInteger("0") } };
    	static BigInteger[][] MOD = {  {  new BigInteger("1") }, {  new BigInteger("1") } };
    	static BigInteger[][] SUM = {  {  new BigInteger("4") }, {  new BigInteger("2") }, {  new BigInteger("1") } };
    
    	private static BigInteger[][] mult(BigInteger[][] cal_fm2, BigInteger[][] mOD2, BigInteger p, boolean flag) { 
    		int i_max = cal_fm2.length;
    		int j_max = mOD2[0].length;
    		int k_max = cal_fm2[0].length;
    		if (k_max != mOD2.length) { 
    			return null;
    		}
    		BigInteger[][] ans = new BigInteger[i_max][j_max];
    		for (int i = 0; i < i_max; i++) { 
    			for (int j = 0; j < j_max; j++) { 
    				BigInteger sum = new BigInteger("0");
    				for (int k = 0; k < k_max; k++) { 
    					if (flag) { 
    						sum = (sum.mod(p)).
    								add(cal_fm2[i][k].multiply(mOD2[k][j]).
    										mod(p)).
    								mod(p);
    					} else { 
    						sum = (sum.add(cal_fm2[i][k].multiply(mOD2[k][j])));
    					}
    				}
    				if (flag) { 
    					ans[i][j] = sum.mod(p);
    				} else { 
    					ans[i][j] = sum;
    				}
    			}
    		}
    		return ans;
    	}
    
    	public static String fib(long n, long m, long p) { 
    		BigInteger mod = new BigInteger("0");
    		BigInteger sum = new BigInteger("0");
    		if (m > n + 2) { 
    			if (n == 1) { 
    				sum = new BigInteger("1");
    			} else { 
    				n = n - 1;
    				while (n != 0) { 
    // System.out.println(n);
    					if ((n & 1) == 1) { 
    						SUM = mult(cal_sum, SUM, new BigInteger(String.valueOf(p)), true);
    					}
    					n = n >> 1;
    					cal_sum = mult(cal_sum, cal_sum, new BigInteger(String.valueOf(p)), true);
    				}
    				sum = SUM[2][0];
    			}
    // System.out.println(sum);
    			return sum.mod(new BigInteger(String.valueOf(p))).toString();
    		} else { 
    			if (m == 1 || m == 2) { 
    				mod = new BigInteger("1");
    			} else { 
    				m = m - 1;
    				while (m != 0) { 
    					if ((m & 1) == 1) { 
    						MOD = mult(cal_fm, MOD, new BigInteger(String.valueOf(p)), false);
    					}
    					m = m >> 1;
    					cal_fm = mult(cal_fm, cal_fm, new BigInteger(String.valueOf(p)), false);
    				}
    				mod = MOD[1][0];
    			}
    			if (n == 1) { 
    				sum = new BigInteger("1");
    			} else { 
    				n = n - 1;
    				while (n != 0) { 
    					if ((n & 1) == 1) { 
    						SUM = mult(cal_sum, SUM, mod, true);
    					}
    					n = n >> 1;
    					cal_sum = mult(cal_sum, cal_sum, mod, true);
    				}
    				sum = SUM[2][0];
    			}
    			return sum.mod(new BigInteger(String.valueOf(p))).toString();
    		}
    	}
    
    	public static void main(String[] args) { 
    		long n, m, p;
    		Scanner scanner = new Scanner(System.in);
    		n = scanner.nextLong();
    		m = scanner.nextLong();
    		p = scanner.nextLong();
    		System.out.println(fib(n, m, p));
    	}
    }

[20200326095424227.png]: /images/20230528/499ccf03e6894d1092c1eeec8ef3a804.png