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

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));}
}