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));
}
}